NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

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# NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

Edited By Sumit Saini | Updated on Aug 18, 2022 02:45 PM IST

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom- In this chapter, we will study about the discovery of fundamental particles like electron, proton and neutrons. NCERT solutions for Class 11 Chemistry chapter 2 Structure of Atom deals with the questions based on various advancements being made to explain different atomic models which gave the basic concept of quantization of energy and gave a modern structure of atoms. In Structure of Atom Class 11 chapter, there are 67 question in the exercise.

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The NCERT solutions for Chemistry Class 11 chapter 2 structure of atom are prepared by our subject experts which provide all the answers to NCERT book questions. These NCERT solutions will help students to clear their doubts in other subjects and other classes as well. By referring to the NCERT solutions for Class 11 , students can understand all the important concepts of NCERT syllabus of Class 11 and practice questions well enough before their examination. Please scroll down to get free NCERT solutions for class 11 chemistry chapter 2 Structure of Atom.

## NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom - Solved Exercise Questions

Question

As the mass of one electron we know is $9.11\times10^{-31}kg$ .

Therefore,

$1g= 10^{-3}kg =\left ( \frac{1}{9.11\times10^{-31}}\right )\times10^{-3}\ electrons$

$\Rightarrow 1.098\times10^{27\ }\ electrons.$ .

Question

As the mass of one electron is equal to $9.11\times10^{-31}kg$

Therefore, Mass of 1 mole or $6.022\times10^{23}$ electrons $= (9.11\times10^{-31})\times(6.022\times10^{23}) = 5.48\times10^{-7}kg$ .

Charge on one electron is $1.602\times10^{-19}\ coulomb$ .

Therefore, the charge on 1 mole of electrons will be:

$\Rightarrow (1.602\times10^{-19})\times(6.022\times10^{23}) = 9.65\times10^{4}\ coulombs$ .

Question

1 molecule of methane $CH_{4}$ contains $6+4 =10$ electrons.

Therefore, 1 mole of methane will contain:

$6.022\times10^{23}\times10 = 6.022\times10^{24}$ electrons.

As 1 atom of 14 C contains $= 14-6= 8\ neutrons$ .

and the number of atoms in 14 C in 1 mole is $6.022\times10^{23}$ atoms.

Therefore, the number of neutrons in 14g of 14 C in 1 mole $=8\times 6.022\times10^{23}\ neutrons$ .

The number of neutrons in $7mg$ :

$= \left ( \frac{8\times6.022\times10^{23}\times7}{14000}\right )$

$= 2.4088\times10^{21}\ neutrons$ .

As the mass of one neutron is $1.674\times10^{-27}kg$ .

Then the mass of total neutrons in 7grams of 14 C:

$= (2.4088\times10^{21})(1.675\times10^{-27}kg) = 4.035\times10^{-6}kg$ .

1 mole of ammonia $NH_{3} =$ $17g\ NH_{3} = 6.022\times10^{23}\ molecules\ of\ NH_{3}$ .

and 1 atom of $NH_{3}$ contains $= 7+3 =10\ protons$ .

Therefore, the number of protons in 1 mole of $NH_{3} = 6.022\times10^{24}\ protons$ .

Number of protons in 3mg of $NH_{3}$ : $=\frac{(6.022\times10^{24}\times34)}{17} \times 1000 = 1.2044\times10^{22}\ protons$

No, there will be no effect of temperature and pressure.

Question

Will the answer change if the temperature and pressure are changed?

As the mass of one proton is $1.6726\times10^{-27}kg$

Therefore, the mass of $1.2044\times10^{22}\ protons$ will be:

$= (1.6726\times10^{-27})\times(1.2044\times10^{22})kg = 2.0145\times10^{-5}kg.$

No, there will be no effect of temperature and pressure.

Question

$^{13}_{\ 6}\textup{C}$

Given the nucleus of carbon: $^{13}_{\ 6}\textup{C}$

Atomic number (Z) = 6

Mass number (A) = 13

Number of protons (Z) = 6

Number of neutrons (A-Z) = 13-6 = 7

Question

$^{16}_{\ 8}\textup{O}$

Given the nucleus of oxygen: $^{16}_{\ 8}\textup{O}$

Atomic number (Z) = 8

Mass number (A) = 16

Number of protons (Z) = 8

Number of neutrons (A-Z) = 16-8 = 8

Question

$_{\ 12}^{\ 24}\textup{Mg}$

Given the nucleus of Magesium: $_{\ 12}^{\ 24}\textup{Mg}$

Atomic number (Z) = 12

Mass number (A) = 24

Number of protons (Z) = 12

Number of neutrons (A-Z) = 24-12 = 12

Question

$^{\ 56}_{\ 26}\textup{Fe}$

Given the nucleus of Iron: $^{\ 56}_{\ 26}\textup{Fe}$

Atomic number (Z) = 26

Mass number (A) = 56

Number of protons (Z) = 26

Number of neutrons (A-Z) = 56-26 = 30

Question

$^{\ 88}_{\ 38}\textup{Sr}$

Given the nucleus of Strontium: $^{\ 88}_{\ 38}\textup{Sr}$

Atomic number (Z) = 38

Mass number (A) = 88

Number of protons (Z) = 38

Number of neutrons (A-Z) = 88-38 = 50

(i) Z = 17 , A = 35.

For the given atomic number Z=17 and mass number A=35;

Atom is $^{35}Cl_{17}$ .

(ii) Z = 92 , A = 233.

For the given atomic number Z=92 and mass number A=233;

Atom is $^{233}U_{92}$ .

(iii) Z = 4 , A = 9.

For the given atomic number Z=4 and mass number A=9;

Atom is $^{9}Be_{4}$ .

Given the wavelength of the yellow light emitted from a sodium, lamp is $\lambda = 580nm = 580\times10^{-9}m$ .

And the frequency will be:

$\nu =\frac{c}{\lambda} = \frac{3.0\times10^8m/s}{580\times10^{-9}m} = 5.17\times10^{14}s^{-1}$

Therefore the wavenumber,

$\vartheta =\frac{1}{\lambda} = \frac{1}{580\times10^{-9}m} = 1.72\times10^{6}m^{-1}$

If a photon has a frequency of $\vartheta = 3\times10^{15} Hz$ .

Then, the energy of each of the photons will be:

$E = h\vartheta = (6.626\times10^{-34}J.s)\times(3\times10^{15}s^{-1})$

$= 1.988\times10^{-18}J$

Question

(ii) have wavelength of 0.50 Å.

For the wavelength $\lambda = 0.50\times10^{-10}m$ .

The energy of each of the photons will be:

$E = h\vartheta = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}J.s)\times(3\times10^{15}s^{-1})}{0.50\times10^{-10}m} = 3.98\times10^{-15}J$

Given frequency, wavelength, and the wave number of a light wave:

$Frequency (\nu ) =\frac{1}{Period} = \frac{1}{2.0\times10^{-10}s}= 5\times10^9 s^{-1}$

$Wavelength (\lambda) = \frac{c}{\nu } = \frac{3.0\times10^8 ms^{-1}}{5\times10^9 s^{-1}} = 6.0\times10^2 m$

$Wave\ number \left ( \vartheta \right ) = \frac{1}{\lambda} = \frac{1}{6.0\times10^2}m = 16.66\ m^{-1}$

Given the wavelength of light $\lambda = 4000\ pm = 4000\times10^{-12}m = 4\times10^{-9} m$ .

and Energy is 1J of energy:

$E = Nh\nu = Nh\frac{c}{\lambda}$

Therefore, the number of photons of light with a wavelength of 4000 pm that provides 1J of energy is:

$N = E\times\frac{\lambda}{hc} = (1J)\times\left ( \frac{4\times10^{-9}m}{(6.626\times10^{-34}J.s)\times(3.0\times10^8 ms^{-1})} \right )$

$= 2.012\times10^{16}\ photons.$

The photon is having a wavelength of $4\times 10^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13\ eV$ .

So, Energy of the photon:

$E = h \nu = \frac{hc}{\lambda}= \frac{(6.626\times10^{-34}J.s)\times (3.0\times10^8 ms^{-1})}{(4\times10^{-7} m)}$

$=4.97\times10^{-19}J$

$=\frac{4.97\times10^{-19}J}{1.602\times10^{-19}}eV$

$= 3.10\ eV$

The photon is having a wavelength of $4\times 10^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13\ eV$ .

The kinetic energy of the emission will be:

$K.E. = h\nu - h\nu_{o} = 3.10 - 2.13\ eV = 0.97\ eV$

The photon is having a wavelength of $4\times 10^{-7}m$ strikes on a metal surface, where the work function of the metal being is $2.13\ eV$ .

From the previous part, we have the Kinetic Energy (K.E.):

$K.E. = \frac{1}{2}mv^2 = 0.97\ eV = 0.97\times1.602\times10^{-19}J$

$\Rightarrow \frac{1}{2}\left ( 9.11\times10^{-31}kg \right )\times v^2 = 0.97\times1.602\times10^{-19}J$

$\because \left ( mass\ of\ an\ electron = 9.11\times10^{-31}kg \right )$

$\Rightarrow v^2 = 0.341\times10^{12} = 34.1\times10^{10}m/s$

$\Rightarrow v = 5.84\times10^5 m/s$

Given the wavelength of the electromagnetic radiation is $242\ nm$ which is just sufficient to ionize the sodium atom.

So, the ionization energy required will be:

$E = Nh\nu = N\frac{hc}{\lambda}$

$= \left ( 6.022\times10^{23} mol^{-1} \right )\left ( \frac{6.022\times10^{23}J.s\times 3.0\times10^8 ms^{-1}}{242\times10^{-9 } m} \right )$

$= 4.945\times10^5 Jmol^{-1} = 494.5 kJmol^{-1}$

Given that the light is monochromatic yellow of wavelength $\lambda = 0.57\mu m$ .

Hence the energy emitted by the bulb will be:

$E = \frac{hc}{\lambda} = \frac{6.626\times10^{-34}Js\times 3.0\times10^{8}m/s}{0.57\times10^{-6} m} = 3.48\times10^{-19 }J$

Therefore, the number of photons emitted per second:

$\frac{25Js^{-1}}{3.48\times10^{-19}J} = 7.18\times10^{19}$

Given the wavelength of radiation is $6800\ \AA$ .

$Energy\ given = Work\ function + Kinetic\ energy.$

But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means the kinetic energy will be zero .

So, the Threshold frequency $\nu_{o}$ will be:

$\nu_{o} = \frac{c}{\lambda_{o}} = \frac{3.0\times10^8 m/s}{6800\times10^{-10}m} = 4.14\times10^{14}s^{-1}$

and the Work function will be:

$W_{o} = h\nu_{o} = 6.626\times10^{-34}Js \times 4.14\times10^{14}s^{-1} = 2.92\times10^{-19}J$

When an electron in a hydrogen atom undergoes a transition from an energy level with $n=4$ to an energy level $n=2$ , there will be an emission of energy whose wavelength can be found by:

$\vartheta =R\left ( \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right )$

$=109677\left ( \frac{1}{2^2}-\frac{1}{4^2} \right )cm^{-1} = 20564.4cm^{-1}$

and wavelength will be equal to:

$\lambda =\frac{1}{\vartheta } = \frac{1}{20564.4} = 486\times10^{-7}cm = 486\times10^{-9}m = 486 nm$

The energy which is required to ionize an H atom if the electron occupies n=5 orbit is:

$E_{n} = \frac{-21.8\times10^{-19}}{n^2} J\ atom^{-1}$

For ionization from 5th orbit, $n_{1} = 5\ and\ n_{2} = \infty$

Therefore, $\triangle E = E_{2}-E_{1} =-21.8\times10^{-19}\times\left ( \frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}\right ) = 21.8\times10^{-19}\times\left ( \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right )$ $=21.8\times10^{-19}\times\left ( \frac{1}{5}^2 - \frac{1}{\infty} \right ) = 8.72\times10^{-20}J$

For ionization from 1st orbit, $n_{1} = 1\ and\ n_{2} = \infty$

Therefore,

$\triangle E' =21.8\times10^{-19}\times\left ( \frac{1}{1}^2 -\frac{1}{\infty} \right ) = 21.8\times10^{-19}J$

$\frac{\triangle E'}{\triangle E} = \frac{21.8\times10^{-19}}{8.72\times10^{-20}} = 25$

Hence, 25 times less energy is required to ionize an electron in the 5th orbital of the hydrogen atom as compared to that in the ground state.

The number of lines produced when an electron from $n^{th}$ shell drops to the ground state:

$= \frac{n(n-1)}{2}$ .

According to the question, the maximum number of emission lines when the excited electron of an H atom in $n=6$ drops to the ground state will be:

$\frac{6(6-1)}{2} = 15$

These are produced because of the following transitions:

$6\rightarrow5$ $6\rightarrow4$ $6\rightarrow3$ $6\rightarrow2$ $6\rightarrow1$

$5\rightarrow4$ $5\rightarrow3$ $5\rightarrow2$ $5\rightarrow1$

$4\rightarrow3$ $4\rightarrow2$ $4\rightarrow1$

$3\rightarrow2$ $3\rightarrow1$

$2\rightarrow1$

The energy associated with the first orbit in the hydrogen atom is $-2.18\times 10^{-18}\ \textup{J atom}^{-1}\ \ \ \ \ \ \ \ \ \ \ \ \(Given)$

The energy of an electron in $n^{th}$ shell is given by:

$E = \frac{-2.18\times10^{-18}}{n^2}J\ atom^{-1}$

So, the energy associated with the fifth orbit would be:

$E = \frac{-2.18\times10^{-18}}{5^2}J\ atom^{-1} = \frac{-2.18\times10^{-18}}{25} J\ atom^{-1}$

$E_{5} = -8.72\times10^{-20} J$

Question

The radius of Bohr's $n^{th}$ orbit for the hydrogen atom is given by,

$r_{n} = (0.0529\ nm)\ n^2$

So, for $n=5$ we have

$r_{5} = (0.0529\ nm) (5)^2$

$r_{5} = 1.3225\ nm$

Balmer formula:

$\vartheta =\frac{1}{\lambda} = R_{H}\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]$

As we can note from the formula that the wavenumber is inversely proportional to the wavelength.

Hence, for the longest wavelength transition in the Balmer series of atomic hydrogen wavenumber has to be the smallest or $n_{2}$ should be minimum i.e., $n_{2} = 3$ .

For the Balmer series, $n_{1} =2$

Thus, the expression of wavenumber is given by,

$\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{1}{2^2}-\frac{1}{3^2} \right ]$

$\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{1}{4}-\frac{1}{9} \right ]$

$\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{9-4}{36} \right ]$

$\vartheta =(1.097\times10^7\ m^{-1})\left [ \frac{5}{36} \right ] = 1.5236\times10^6\ m^{-1}$

The ground state energy:

$E_{1} = -2.18\times10^{-11}\ ergs$

$= -2.18\times10^{-11}\times10^{-7} J$

$= -2.18\times10^{-18} J$

The energy required to shift the electron from the first Bohr orbit to the fifth Bohr orbit is:

$\triangle E = E_{5}-E_{1}$

And the expression for the energy of an electron is given by:

$E_{n} =-\frac{2n^2me^4Z^2}{n^2h^2}$

where m is mass of an electron, Z is the atomic mass of an atom, e is a charge of an electron, and h is the Planck's constant.

Now, substituting the values in the equation, we get

$=-\frac{(2.18\times10^{-18})(1)^2}{(5)^2} - (-2.18\times10^{-18})$

$=(2.18\times10^{-18})\left [ 1-\frac{1}{25} \right ]$

$=(2.18\times10^{-18}) \left ( \frac{24}{25}\right ) = 2.0928\times10^{-18}J$

Hence, the wavelength of the emitted light will be:

$\lambda = \frac{hc}{E} = \frac{(6.626\times10^{-34}Js)(3\times10^8m/s)}{(2.0928\times10^{-18}J)}$

$= 9.498\times10^{-8}\ m$

The expression for the energy of an electron in hydrogen is:

$E_{n}=-\frac{2n^2me^4Z^2}{n^2h^2}$

Where m is mass of electrons, Z is the atomic mass of an atom, e is the charge of an electron, and h is the Planck's constant.

and electron energy in the hydrogen atom is given by,

$E_{n}=-\frac{(2.18\times10^{-18})}{n^2}J$

The electron energy in $n=2$ orbit is:

$E_{n}=-\frac{(2.18\times10^{-18})}{2^2}J = 0.5465\times10^{-18}J$

Therefore, the energy required for the ionization from $n=2$ is $5.45\times10^{-19}J$

Now, the longest wavelength of light that can be used to cause this transition will be:

$E=\frac{hc}{\lambda}$

$\lambda = \frac{hc}{E} =\frac{(6.626\times10^{-34}Js)(3\times10^8)m/s}{5.45\times10^{-19}J}$

$=3.674\times10^{-7}m = 3.674\times10^{-5}cm$

The wavelength of an electron is given by the de Broglie's equation:

$\lambda = \frac{h}{mv}$

Where,

$\lambda$ is the wavelength of moving particle,

m is the mass of the particle, i.e., $9.11\times10^{-31}kg$

v is the velocity of the particle,i.e., $2.05\times 10^7\ \textup{ms}^{-1}$ ( Given)

and h is the Planck's constant value, i.e., $(6.626\times10^{-34}Js)$

Now, substituting the values in the equation, we get

$\lambda = \frac{(6.626\times10^{-34}Js)}{(9.11\times10^{-31}kg)(2.05\times10^{7}m/s)} = 3.548\times10^{-11}m$

Hence, the wavelength of the electron moving with a velocity of $2.05\times 10^7\ \textup{ms}^{-1}$ is $3.548\times10^{-11}m$ .

The wavelength of an electron can be found by de Broglie's equation:

$\lambda = \frac{h}{mv}$

Given the K.E. of electron $3.0\times 10^{-25}\ \textup{J}$ which is equal to $\frac{1}{2}mv^2$ .

Hence we get,

$\frac{1}{2}mv^2 = 3.0\times10^{-25}J$

$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(3.0\times10^{-25}J)}{9.1\times10^{-31}kg}} = 811.579\ m/s$

Hence the wavelength is given by,

$\lambda = \frac{6.626\times10^{-34}Js}{(9.1\times10^{-31}kg)(811.579\ m/s)}$

$= 8.9625\times10^{-7}m$

$\textup{Na}^{+},\ \textup{K}^{+}, \ \textup{Mg}^{2+}, \ \textup{Ca}^{2+}, \ \textup{S}^{2-},\ \textup{Ar}$

Calculating the number of electrons for each species.

$Na$ has 11 electrons then, $Na^{+}$ will have $(11-1) =10$ electrons.

$K$ has 19 electrons then, $K^{+}$ will have $(19-1) =18$ electrons.

$Mg$ has 12 electrons then, $Mg^{2+}$ will have $(12-2) =10$ electrons.

$Ca$ has 20 electrons then, $Ca^{2+}$ will have $(20-2) =18$ electrons.

$S$ has 16 electrons then, $S^{2-}$ will have $(16+2) =18$ electrons.

$Ar$ has 18 electrons.

Hence, the following are isoelectronic species:

$Na^{+}\ and\ Mg^{2+}$ having 10 electrons each.

$K^{+},\ Ca^{2+}, S^{2-}, and\ Ar$ having 18 electrons each.

Question :

(a) $\textup{H}^-$

The electronic configuration of $H$ is : $1s^1$

Now, the electronic configuration of $\textup{H}^-$ will be $1s^2$ .

Question

(b) $\textup{Na}^+$

The electronic configuration of $Na$ having $Z=11$ is : $1s^22s^22p^63s^1$

Now, the electronic configuration of $\textup{Na}^+$ will be $1s^22s^22p^6$ .

Question

(c) $\textup{O}^{2-}$

The electronic configuration of $O$ having $Z=8$ is : $1s^22s^22p^4$

Now, the electronic configuration of $\textup{O}^{2-}$ will be $1s^22s^22p^6$ .

(a) $\textup{3s}^1$

With given outermost electrons $\textup{3s}^1$ ,

The complete electronic configuration is $1s^22s^22p^63s^1$

Hence the number of electrons present in the atom of the element is:

$= 2+2+6+1 = 11$ .

Therefore, the atomic number of the element is $11$ which is Sodium (Na) .

(b) $2p^3$

With given outermost electrons $2p^3$ ,

The complete electronic configuration is $1s^22s^22p^3$

Hence the number of electrons present in the atom of the element is:

$= 2+2+3= 7$ .

Therefore, the atomic number of the element is 7 which is NItrogen (N) .

(c) $3p^5$

With given outermost electrons $3p^5$ ,

The complete electronic configuration is $1s^22s^22p^63s^23p^5$

Hence the number of electrons present in the atom of the element is:

$= 2+2+6+2+5= 17$ .

Therefore, the atomic number of the element is 17 which is Chlorine (Cl) .

Question

(a) $\left [\textup{He} \right ]2s^1$

The electronic configuration of the element is $\left [\textup{He} \right ]2s^1$ or $1s^22s^1.$

Therefore, the atomic number of the element is 3 which is Lithium , a p-block element.

Question

(b) $[\textup{Ne}]3s^23p^3$

The electronic configuration of the element is $[\textup{Ne}]3s^23p^3$ or $1s^22s^22p^63s^23p^3.$

Therefore, the atomic number of the element is 15 which is Phosphorus , a p-block element.

Question

(c) $[Ar]4s^2\ 3d^1$

The electronic configuration of the element is $[Ar]4s^2\ 3d^1$ or $1s^22s^22p^63s^23p^64s^23d^1.$

Therefore, the atomic number of the element is 21 which is Scandium , a d-block element.

For g-orbital, the value of Azimuthal quantum number (l) will be 4.

As for any value 'n' of the principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n-1).

Therefore, for l =4, the minimum value of n should be 5.

For d-orbital, the value of Azimuthal quantum number (l) = 2.

When l =2, the values of m are: $-2,-1,0,+1,+2$

Now, for the 3d orbital:

The value of Principal quantum number, $n = 3$

Azimuthal quantum number, $l = 2$

Magnetic quantum number, $m_{l}$ = $-2,-1,0,1,2$

Given an atom of an element contains 29 electrons and 35 neutrons.

Now, for an atom to be neutral, the number of protons is equal to the number of electrons.

Therefore, the number of protons in the atom of the given element will be 29 .

Given an atom of an element contains 29 electrons and 35 neutrons.

The electronic configuration of the atom will be:

$1s^22s^22p^63s^23p^64s^13d^{10}$ which is the electronic configuration of copper .

The number of electrons in $\textup{H}_2$ molecule is $(1+1) =2$ .

The number of electrons in $\textup{H}_2^+$ molecule will be one less than the number of electrons in $H_{2}$ molecule. i.e, $(2-1 )=1$ .

The number of electrons present in $\textup{O}_2^+$ molecule will be one less than the number of electrons present in $O_{2}$ molecule. i.e., $(16-1) =15$

For a given value of the principal quantum number $(n)$ , the azimuthal quantum number $(l)$ can have values from $0$ to $(n-1)$ .

Therefore, for given atomic orbital $n =3$ ,

The value of $l$ can take values from $0$ to $(3-1) =2$ , i.e., $l = 0,1,2$ .

And for a given value of $l$ , the Magnetic quantum number $m_{l}$ can have $(2l+1)$ values.

When value of $l =0$ then, $m=0$ ,

or $l=1$ then, $m= -1,0,+1$

or $l=2$ then, $m= -2,-1,0,+1,+2$

or $l=3$ then, $m= -3,-2,-1,0,+1,+2,+3$

Question

For 3d-orbital, the values of Principal quantum number is $(n) =3$ and Azimuthal quantum number $(l) = 2$ .

Therefore, for $(l) = 2$ ,

$m_{l}$ , magnetic quantum number can have $(2l+1)$ values.

i.e., $m_{l} = -2,-1,0,+1,+2$

Question

$1p$ is NOT possible because for $n=1$ , the value of $l$ is zero. $(for \ p,l=1)$

$2s$ is possible because, when $n=2,l=0,1$ . $(for \ s,l=0)$ .

$2p$ is possible because when $n=2,l=0,1$ . $(for \ p,l=1)$ .

$3f$ is NOT possible because for $n=3$ , the value of $l=0,1,2$ . $(for \ f,l=3)$

(a) $n = 1, l = 0$

Here, $n$ is principal quantum number and $l$ is azimuthal quantum number.

Then the orbital with given quantum numbers $n = 1, l = 0$ is $1s$ which can have a maximum of 2 electrons.

(b) $n = 3, l = 1$

Here, $n$ is principal quantum number and $l$ is azimuthal quantum number.

Then the orbital with given quantum numbers $n = 3, l = 1$ is $3p$ which can have a maximum of 6 electrons.

(c) $n = 4, l = 2$

Here, $n$ is principal quantum number and $l$ is azimuthal quantum number.

Then the orbital with given quantum numbers $n = 4, l = 2$ is $4d$ which can have a maximum of 10 electrons.

(d) $n = 4, l = 3$

Here, $n$ is principal quantum number and $l$ is azimuthal quantum number.

Then the orbital with given quantum numbers $n = 4, l = 3$ is $4f$ which can have a maximum of 14 electrons.

$n=0,l=0\: \: \: \: \: m_{1}=0,m_{s}=+1/2$

Given quantum numbers : $n=0,l=0\: \: \: \: \: m_{1}=0,m_{s}=+1/2$

NOT possible, because n cannot be equal to zero.

$n=1,l=0\: \: \: \: \: m_{1}=0,m_{s}=-1/2$

Given quantum numbers : $n=1,l=0\: \: \: \: \: m_{1}=0,m_{s}=-1/2$

It is possible and it is 1s orbital.

$n=1,l=1\: \: \: \: \: m_{1}=0,m_{s}=+1/2$

Given quantum numbers : $n=1,l=1\: \: \: \: \: m_{1}=0,m_{s}=+1/2$

It is NOT possible because when $n=1,\ l\neq 1$ .

$n=2,l=1\: \: \: \: \: m_{1}=0,m_{s}=-1/2$

Given quantum numbers :

$n=2,l=1\: \: \: \: \: m_{1}=0,m_{s}=-1/2$

It is possible and it is 2p orbital.

$n=3,l=3\: \: \: \: \: m_{1}=-3,m_{s}=+1/2$

Given quantum numbers :

$n=3,l=3\: \: \: \: \: m_{1}=-3,m_{s}=+1/2$

It is NOT possible because when $n=3$ , $l \neq 3$ .

$n=3,l=1\: \: \: \: \: m_{l}=0,m_{s}=+1/2$

Given quantum numbers :

$n=3,l=1\: \: \: \: \: m_{l}=0,m_{s}=+1/2$

It is possible and it is 3p orbital.

(a) $n = 4, m_s = -^{\ 1}/_2$

The total number of electrons in an atom for a value of n is given by: $n=2n^2$

Therefore, the total no. of electrons when $n=4$ ,

$\Rightarrow n=2\times4^2 = 32$ and half of them i.e. 16 will have $m_{s} = -\frac{1}{2}$ .

(b) $n =3, \ l = 1$

When $n =3\ and\ l = 1$ then it is $3s$ orbital which can have 2 electrons .

According to the Bohr's postulate of angular momentum,

$mvr=\frac{nh}{2\pi}$

which can be written as: $2\pi r=\frac{nh}{mv}$ .....................................(1)

Then according to de Broglie's equation for wavelength,

$\lambda = \frac{h}{mv}$ ......................................(2)

Now, substituting the values of equation (2) in equation (1) we get,

$2\pi r = n\lambda$

Thus, the circumference of the Bohr's orbit for the hydrogen atom is an integral multiple of de Broglie's wavelength associated with the electron revolving around the orbit.

For the transition of H-like particles,

$\vartheta = \frac{1}{\lambda} = R_{H}Z^2\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]$

For $\textup{He}^+$ transition spectrum,

$Z=2$ , $n_{2}=4$ , and $n_{1}=2$

Therefore,

$\vartheta = \frac{1}{\lambda} = R_{H}2^2\left [ \frac{1}{2^2}-\frac{1}{4^2} \right ] = \frac{3R_{H}}{4}$

Then for the hydrogen spectrum,

$\vartheta = \frac{3R_{H}}{4}$ and $Z=1$

Therefore,

$\vartheta = \frac{1}{\lambda} = R_{H}\times1\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ]$

$\Rightarrow R_{H}\times1\left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ] = \frac{3R_{H}}{4}$

$\Rightarrow \left [ \frac{1}{n_{1}^2}-\frac{1}{n_{2}^2} \right ] = \frac{3}{4}$

The values of $n_{1}\ and\ n_{2}$ can be found by the hit and trial method in the above equation.

So, we get $n_{1} = 1$ and $n_{2} = 2$ , i.e., the transition is from $n =2$ to $n =1$ .

Question

$\textup{He}^+(g) \rightarrow \textup{He}^{2+} (g) + \textup{e}^-$

The ionization energy for the H atom in the ground state is $2.18 \times 10^{-18}\ \textup{J atom}^{-1}$

For the hydrogen-like particles,

$E_{n} =-\frac{2n^2mZ^2e^4}{n^2h^2}$

For H-atom, Ionization energy:

$I.E. =E-E_{1} = 0-(-\frac{2\pi^2me^4}{1^2h^2})= \frac{2\pi^2me^4}{h^2}$

$= 2.18\times10^{-18}J/atom$ $(Given)$

For the given process, the energy required will be:

$E_{n} - E_{1}$

$= 0-(-\frac{2\pi^2me^4}{1^2h^2})$

$= 4\times\frac{2\pi^2me^4}{h^2}$

$= 4\times2.18\times10^{-18}J$

$= 8.72\times10^{-18}J$

Given the diameter of a carbon atom which is $0.15\ nm = 0.15\times10^{-9}m = 1.5\times10^{-10} m$

Then the number of carbon atoms which can be placed side by side in a straight line across the length of the scale of length $20\ cm$ long will be:

$=\frac{2\times10^{-1}m}{1.5\times10^{-10}m} = 1.33\times10^{9}$

The arrangement length is given which is $2.4\ cm = 2.4\times10^{-2} m$ .

and the number of atoms of carbon which are arranged in this length is given $2\times10^8$ .

Let the radius of carbon atom be $'r'$ then,

$(2r)\times(2\times10^8) = 2.4\times10^{-2} m$

$\Rightarrow r = \frac{2.4\times10^{-2}m}{(2\times10^8)\times2} = 0.060\times10^{-9}m$ or $r = 0.060\ nm$

Hence the radius of carbon atom is $r = 0.060\ nm$ .

If the diameter of zinc atom is $2.6\ \AA$ then, its radius would be:

$= \frac{2.6\ \AA}{2} = 1.3\ \AA = 1.3\times10^{-10} m\ or\ 130\ nm$

The number of atoms present in a length of $1.6\ cm$ will be:

$= \frac{1.6\times10^{-2}m}{2.6\times10^{-10}m} = 0.6153\times10^8m$

$= 6.153\times10^7$

As the charge carried by one electron is $1.602\times10^{-19}C$

Therefore, the number of electrons present in particle carrying $2.5\times 10^{-16}\ \textup{C}$ charge will be:

$=\frac{2.5\times10^{-16}C}{1.6022\times10^{-19}C} = 1.560\times10^3$

$= 1560\ electrons$

Given charge on the oil drop is $-1.282\times10^{-18}C$

and the charge carried by one electron is $-1.6022\times10^{-19} C$

Therefore, the number of electrons present on the oil drop carrying $-1.282\times10^{-18}C$ charge is:

$= \frac{-1.282\times10^{-18}C}{-1.6022\times10^{-19}C}$

$= 0.8001\times10 = 8.0$

The thin foil of heavy atoms, like gold, platinum, etc. have a nucleus carrying a large amount of positive charge. Therefore, some $\alpha$ particles will easily get deflected back.

These $\alpha$ particles also deflect through small angles because of the large number of a positive charge.

Hence if we use light atoms, their nuclei will have a small positive charge, hence the number of $\alpha$ particles getting deflected even through small angles will be negligible.

The general way to represent an element along with its atomic mass (A) and atomic number (Z) is $_{Z}^{A}\textrm{X}$ . Here the atomic number of an element is fixed. However, its mass number is not fixed as it depends upon the isotope taken.

Hence, it is essential to indicate the mass number.

Let the number of protons of an atom be $'p'$ .

Then the number of neutrons will be,

$p + \frac{31.7}{100}p = 1.317p$

and the mass number is $81$ (Given).

Mass number = number of neutrons + number of protons.

Therefore,

$81 = 1.317p + p$

$\Rightarrow p = \frac{81}{2.317} =35$

Thus, there are $35$ numbers of protons which is also its atomic number.

Hence, the symbol for the element is $_{35}^{81}\textrm{Br}$ .

Given an ion has mass number 37 and possesses one unit of negative charge.

Let the number of electrons be $'e'$ then,

the number of neutrons will be: $e+\frac{11.1e}{100} = 1.111e$

The number of electrons in the neutral atom $= e-1$ (ion possesses one unit of negative charge).

Therefore, the number of protons will be $= e-1$ .

Mass number = number of protons + number of neutrons

therefore,

$1.111e +e -1 =37$

$\Rightarrow 2.111e =38$

$\Rightarrow e =18$

Therefore, the number of protons is equal to the atomic number.

$\Rightarrow e-1 = 18-1 =17$

Hence, the symbol for an ion will be:

$_{17}^{37}\textrm{Cl}^{-1}$

Given an ion has mass number 56 and possesses three units of negative charge.

Let the number of electrons be $'e'$ then,

the number of neutrons will be: $e+\frac{30.4e}{100} = 1.304e$

The number of electrons in the neutral atom $= e+3$ (ion possesses three units of positive charge).

Therefore, the number of protons will be $= e+3$ .

Mass number = number of protons + number of neutrons

therefore,

$1.304e +e +3 =56$

$\Rightarrow 2.304e =53$

$\Rightarrow e =23$

Therefore, the number of protons is equal to the atomic number.

$\Rightarrow e+3 = 23+3 =26$

Hence, the symbol for an ion will be:

$_{26}^{56}\textrm{Fe}^{3+}$

The increasing order of frequency of radiations will be:

Radiation from FM radio < amber light from traffic signal < radiation from microwave oven < X-rays < cosmic rays from outer space.

The energy emitted by the nitrogen laser is:

$E = Nh\nu = N\frac{hc}{\lambda} = \left ( 5.6\times10^{24} \right )\times\frac{(6.626\times10^{-34}Js)(3.0\times10^{8}m/s)}{337.1\times10^{-9}m}$

$=3.3\times10^6J$

The wavelength of neon gas is $616\ nm$ or $616\times10^{-9}m$

Hence the frequency of this radiation will be:

$\nu = \frac{c}{\lambda} = \frac{3.0\times10^8m/s}{616\times10^{-9}m} = 4.87\times10^{14}s^{-1}$

The velocity of neon gas radiation is $3.0\times10^8m/s$ .

Therefore, the distance travelled in $30s$ will be:

$30s\times(3.0\times10^8m/s) = 9.0\times10^9m$

The energy of quantum will be:

$E = h\nu = h\frac{hc}{\lambda}$

$= \frac{(6.626\times10^{-34}Js)\times(3.0\times10^8m/s)}{616\times10^{-9}m} = 32.27\times10^{-20}J$

If it produces 2J of energy then, the number of quanta present in it will be $'N'$ .

Therefore,

$E=Nh\nu$ or $N=\frac{E}{h\nu}$ ,

Where $E = 2J$ and $h\nu = 32.27\times10^{-20}J$ from the previous part.

$N=\frac{2J}{32.27\times10^{-20}J} = 6.2\times10^{18}$

Let the number of photons received by the detector be $'N'$ .

Then, the total energy it receives from the radiation of $600 nm$ will be:

$E=Nh\nu = N\frac{hc}{\lambda}$ Or $N = \frac{\lambda E}{hc}$

Where,

$\lambda = 600\ nm = 600\times10^{-9}m$ ,

$E = 3.15\times10^{-18}J$ ,

$h = 6.626\times10^{-34}Js$ and $c = 3.0\times10^8 m/s$

Substituting the values in the equation above, we get

$N = \frac{(3.15\times10^{-18}J)\times(600\times10^{-9}m)}{(6.626\times10^{-34}Js)\times(3.0\times10^8m/s)}$

$= 9.51 \approx 10$ number of photons is received by the detector.

Given the duration of a radiation source $= 2\ ns = 2\times10^{-9}s$ and the number of pulse source is $= 2.5\times10^{15}$ , then its frequency will be:

$\nu = \frac{1}{2\times10^{-9}s} = 0.5\times10^{9}s^{-1}$

and the energy of the source for the given frequency will be:

$E= Nh\nu = (2.5\times10^{15})(6.626\times10^{-34}Js)\times(0.5\times10^{9}s^{-1})$

$= 8.28\times10^{-10}J$

We have two wavelengths of $\lambda_{1} = 589\ nm= 589\times10^{-9}m$ and $\lambda_{2} = 589.6\ nm = 589.6\times10^{-9}m$ .

Calculating the frequency for each:

$\nu_{1} = \frac{c}{\lambda_{1}} = \frac{3.0\times10^{8}m/s}{589\times10^{-9}m} = 5.093\times10^{14}s^{-1}$

$\nu_{2} = \frac{c}{\lambda_{2}} = \frac{3.0\times10^{8}m/s}{589.6\times10^{-9}m} = 5.088\times10^{14}s^{-1}$

Therefore, the energy difference between two excited states will be:

$\triangle E =E_{2}-E_{1} = h(\nu_{2}-\nu_{1})$

$= (6.626\times10^{-34}Js)(5.093-5.088)\times10^{14}s^{-1}$

$= 3.31\times10^{-22}J$

Given the work function for the Caesium atom is $1.9\ eV$ .

i.e., $W_{o} = h\nu_{o} = \frac{hc}{\lambda_{o}}$ Or $\lambda_{o} = \frac{hc}{W_{o}}$

As $1\ eV = 1.602\times10^{-19}J$

$\therefore 1.9\ eV =1.9\times1.602\times10^{-19}J =3.0438\times10^{-19}J$

$\lambda_{o} = \frac{(6.626\times10^{-34}Js)(3.0\times10^8)J}{3.0438\times10^{-19}J} =6.5306\times 10^{-7} m$

Therefore, the threshold wavelength is 6.53× 10 -7 m.

To find threshold frequency:

$W_{o} = h\nu_{o}$

Where

h = Planck’s constant

$\nu_{o}$ = threshold frequency

$\nu_{o}= \frac{W_{o}}{h}$

$\nu_{o} =\frac{ [1.9\times1.602\times10^{-19}J] }{[6.626\times10^{-34}Js]}$

$= \frac {3.0438\times10^{-19}J} { 6.626\times10^{-34}Js}$

$= 4.5937\times10^{14} Hz$

Finding the kinetic energy of the ejected electrons:

K.E of the ejected photoelectron:

$= h[\nu-\nu_{o}] = hc\left [\frac{1}{\lambda} -\frac{1}{\lambda_{o}} \right ]$

$=(6.626\times10^{-34}Js)(3.0\times10^8m/s)\left [ \frac{1}{500\times10^{-9}m} - \frac{1}{654\times10^{-9}m} \right ]$

$= 19.878\times10^{-26}Jm\left [ \frac{154}{327000}\times10^9 m^{-1}\right ]$

$= 9.361\times10^{-20}J$

Finding the Velocity of the ejected electrons:

$KE = \frac{1}{2}mv^2$

Where,

$m$ = mass of electron

$v$ = velocity of electron

Therefore, the velocity is given by,

$v =\sqrt{ \frac{2KE}{m}}$

$v = \sqrt{\frac{2\times9.361\times10^{-20}J}{9.1\times10^{-31}kg}} = 4.52\times10^5\ m/s$

$\\\lambda\ (\textup{nm})\qquad \qquad \qquad 500\quad 450\quad 400\\ \textup{v}\times 10^{-5} \ (\textup{cm s}^{-1})\quad \; 2.55\ \ 4.35\ \ 5.35$

Let us assume the threshold wavelength to be $\lambda\ nm$ and the kinetic energy of the radiation is given as:

$h(\nu-\nu_{o}) = \frac{1}{2}mv^2$

$hc\left ( \frac{1}{\lambda} - \frac{1}{\lambda_{o}} \right )= \frac{1}{2}mv^2$

$\Rightarrow hc\left ( \frac{1}{500\times10^9} - \frac{1}{\lambda\times10^{-9}} \right ) = \frac{1}{2}m\left ( 2.55\times10^5\times10^{-2} \right )^2$

$\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{500}-\frac{1}{\lambda} \right )= \frac{1}{2}m(2.55\times10^3)^2$ .................................(1)

Similarly, we can also write,

$\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{450}-\frac{1}{\lambda} \right )= \frac{1}{2}m(3.45\times10^3)^2$ ...................................(2)

$\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{400}-\frac{1}{\lambda} \right )= \frac{1}{2}m(5.35\times10^3)^2$ ...................................(3)

Now, dividing the equations (3) with (1),

$\frac{\left ( \frac{\lambda -400}{400\lambda} \right )}{\left (\frac{\lambda -500}{500\lambda} \right )} = \frac{\left ( 5.35\times10^3 \right )^2}{\left ( 2.55\times10^3 \right )^2}$

$\frac{5\lambda -2000}{4\lambda -2000} = \frac{\left ( 5.35 \right )^2}{\left ( 2.55 \right )^2}$

$\frac{5\lambda -2000}{4\lambda -2000} = 4.40177$

$\Rightarrow 17.6070\lambda -5\lambda = 8803.537 -2000$

$\Rightarrow \lambda = \frac{6805.537}{12.607} = 539.8\ nm$

Therefore, the wavelength is $540\ nm$ .

$\\\lambda\ (\textup{nm})\qquad \qquad \qquad 500\quad 450\quad 400\\ \textup{v}\times 10^{-5} \ (\textup{cm s}^{-1})\quad \; 2.55\ \ 4.35\ \ 5.35$

We have the threshold wavelength. $\lambda_{o} = 540\ nm$

Then substituting this value in any of the equation ( look in the previous part), we get

$\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{400}-\frac{1}{\lambda} \right )= \frac{1}{2}m(5.35\times10^3)^2$

$\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{400}-\frac{1}{540} \right )= \frac{1}{2}m(5.35\times10^3)^2$

Taking the mass of an electron to be $9.11\times10^{-31}kg$ .

$h= \frac{(9.11\times10^{-31})(5.35\times10^3)^2\times10^{-9}\times(400\times540)}{2\times(3.0\times10^8)(140)}$

$= 6.705\times10^{-34} Js$ approximately.

Given work function of the metal, $W$ and the Wavelength, $\lambda = 256.7\ nm$

From the Law of conservation of energy, the energy of an incident photon E is equal to the sum of the work function W of radiation and its kinetic energy K.E i.e.,

$E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}Js)(3.0\times10^8)m/s}{256.7\times10^{-9}m}$

$= 7.74\times10^{-19} J$ Or

$= \frac{7.74\times10^{-19}J}{1.602\times10^{-19}}eV = 4.83\ eV$

Since the potential applied gives the kinetic energy to the radiation, therefore K.E of the electron $= 0.35\ eV$

Therefore, Work Function $W$ $= 4.83 - 0.35 = 4.48\ eV$

Given the wavelength of a photon which strikes an atom is $150\ pm = 150\times10^{-12} m$ .

Then the energy associated with this photon will be:

$E= \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}Js)(3.0\times10^8)m/s}{150\times10^{-12}m}$

$= 1.3252\times10^{-15}J$

Given the velocity of ejected inner bounded electron: $v= 1.5\times10^7 m/s$ .

Then, the energy associated with this electron will be, Kinetic energy.

Hence finding

$KE = \frac{1}{2}mv^2$

Where

m = mass of electron, v = velocity of electron

$KE = \frac{1}{2}(9.11\times10^{-31}kg)(1.5\times10^7)^2$

$= 1.02\times10^{-16}J$

Hence the energy with which the electrons are bounded to the nucleus is:

$= 13.25\times10^{-16}J - 1.02\times10^{16}J$

$= 12.23\times10^{-16}J$

$=\frac{ 12.23\times10^{-16}J}{1.602\times10^{-19}}$

$= 7.63\times10^3\ eV$

Given transition in the Paschen series end at orbit $n=3$ and starts from orbit n: $v = 3.29\times 10^{15}\ (\textup{Hz})\left[\frac{1}{3^2} - \frac{1}{n^2} \right ]$ .........................(1)

$\nu = \frac{c}{\lambda} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m}$ .........................(2)

Equating both (1) and (2) equations: we get

$3.29\times10^{15}\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m}$

$\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m\times3.29\times10^{15}}$

$\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = 0.07096 s^{-1}$

$\frac{1}{9} - 0.07096 = \frac{1}{n^2}$

$\frac{1}{n^2} = 0.0401 \Rightarrow n^2 =25 \Rightarrow n=5$

Therefore, the radiation corresponding to 1285 nm lies in the infrared region .

The radius of $n^{th}$ orbit of H-like particles is given by:

$=0.529\times\frac{n^2}{Z} \AA$ Or $=52.9\times\frac{n^2}{Z}\ pm$

Here, starting radius, $r_{1} = 1.3225\ nm = 1322.5\ pm = 52.9n_{1}^2$

Ending radius, $r_{2} = 211.6\ pm = 52.9\left (\frac{n_{2}^2}{Z} \right )$

Therefore, $\frac{r_{1}}{r_{2}} = \frac{1322.5\ pm}{211.6\ pm} = \frac{n_{1}^2}{n_{2}^2}$

$\Rightarrow \frac{n_{1}^2}{n_{2}^2} = 6.25$

$\Rightarrow \frac{n_{1}}{n_{2}} = 2.5$

If $n_{2} = 2$ and $n_{1} = 5$ , then the transition is from $5^{th}$ orbit to $2^{nd}$ orbit.

Therefore, it belongs to the Balmer Series .

Frequency $\nu$ is given by:

$= 1.097\times10^7 m^{-1}\left ( \frac{1}{2^2}-\frac{1}{5^2} \right )$

$= 1.097\times10^7\left (\frac{21}{100} \right )m^{-1}$

Wavelength :

$\lambda = \frac{1}{\nu} = \frac{100}{1.097\times21\times10^7}m = 434\times10^{-9}m = 434\ nm$ .

Therefore, it lies in the visible range.

According to de- Broglie's equation for the wavelength.

$\lambda = \frac{h}{mv}$

Given the velocity of electron $v = 1.6\times10^{6} m/s$

and mass of electron $m = 9.11\times10^{-31}kg$

So, the wavelength will be:

$\lambda = \frac{(6.626\times10^{-34}Js)}{(9.11\times10^{-31}kg)(1.6\times10^6 m/s)}$

$= 4.55\times10^{-10}m$ or $455\ pm$

Given the wavelength of neutron: $\lambda = 800\ pm$

and the mass of neutron $m =1.675\times10^{-27}kg$

So, According to the de-Broglie's equation,

$\lambda = \frac{h}{mv}$

Substituting the values in above equation:

$800\times10^{-12} m = \frac{6.626\times10^{-34}Js}{(1.675\times10^{-27}kg)\times(v)}$

$\Rightarrow v = \frac{6.626\times10^{-34}Js}{(1.675\times10^{-27}kg)\times(800\times10^{-12}m)}$

$v= 4.94\times10^4 m/s$

Given the velocity of the electron in Bohr's first orbit is $2.19\times 10^6 \ \textup{ms}^{-1}$ .

And we know the mass of electron which is $9.11\times10^{-31}kg$

Hence the de-Broglie's wavelength associated with the electron will be:

$\lambda = \frac{h}{mv} = \frac{6.626\times10^{-34}Js}{(9.11\times10^{-31}kg)\times(2.19\times10^6m/s)}$

$= 3.32\times10^{-10}m = 332\ pm$

Given a proton is moving with velocity $4.37\times 10^5\ \textup{ms}^{-1}$ .

And if the hockey ball of mass $0.1kg$ is also moving with the same velocity, then

According to de-Broglie's equation we have,

$\lambda = \frac{h}{mv}$

$\lambda =\frac{6.626\times10^{-34}Js}{(0.1kg)\times(4.37\times10^5 m/s)}$

$= 1.516\times10^{-28} m$

We have given the uncertainty in position, i.e., $\triangle x= \pm 0.002\ nm = 2\times10^{-12} m$ .

According to Heisenberg's Uncertainty Principle:

$\triangle x\times \triangle p = \frac{h}{4\pi}$

Where,

$\triangle x$ is uncertainty in the position of the electron.

$\triangle p$ is uncertainty in the momentum of the electron.

Then, $\triangle p = \frac{h}{4\pi\times\triangle x}$

$\triangle p =\frac{6.626\times10^{-34}Js}{4\pi\times(2\times10^{-12}m)} =2.636\times10^{-23}Jsm^{-1}$

Or $2.636\times10^{-23}kgms^{-1}$ $\left ( 1J- 1kgms^2s^{-1} \right )$

The actual momentum of the electron:

$\frac{h}{4\pi_{m}\times0.05\ nm}= \frac{6.626\times10^{-34}Js}{4\pi\times0.05\times10^{-9} m }$

$\Rightarrow p = 1.055\times10^{-24}kg\ m/s$

Therefore, it cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.

$\\1.\quad n = 4, \ l =2,\ m_l = -2,\ m_s = -1/2 \\2.\quad n = 3, \ l =2,\ m_l = 1,\ m_s = +1/2 \\3.\quad n = 4, \ l =1,\ m_l = 0,\ m_s = +1/2 \\4.\quad n = 3, \ l =2,\ m_l = -2,\ m_s = -1/2 \\5.\quad n = 3, \ l =1,\ m_l = -1,\ m_s = +1/2 \\6.\quad n = 4, \ l =1,\ m_l = 0,\ m_s = +1/2$

Quantum number provides the entire information about an electron of a particular atom.

Principal quantum number $'n'$

Azimuthal quantum number $'l'$

Magnetic quantum number $'m_{l}'$

Spin quantum number $'m_{s}'$ .

The orbitals occupied by the electrons are:

1. 4d-orbital

2. 3d-orbital

3. 4p-orbital

4. 3d-orbital

5. 3p-orbital

6. 4p-orbital

For the same orbitals, electrons will have the same energy and higher the value of $(n+l)$ value higher is the energy.

Therefore, the increasing order of energies:

$5.{\color{DarkGreen} (3p)}< 2.{\color{DarkGreen} (3d)} = 4.{\color{DarkGreen} (3d)}<3.{\color{DarkGreen} (4p)}=6. {\color{DarkGreen} (4p)}<1.{\color{DarkGreen} (4d)}$

As the p-orbital is farthest from the nucleus hence the electrons in (4p)subshell experiences the lowest effective nuclear charge.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, (i) 2s and 3s

The 2s orbital is more closer to the nucleus than 3s orbital hence 2s will experience a larger effective nuclear charge compared to 3s.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, (ii) 4d and 4f

The 4d orbital is more closer to the nucleus than 4f orbital hence 4d will experience a larger effective nuclear charge compared to 4f.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Closer orbitals experience more nuclear charge than outer orbitals.

Therefore, (ii) 3d and 3p

The 3p orbital is more closer to the nucleus than 3d orbital hence 3p will experience a larger effective nuclear charge compared to 3d.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.

Silicon has a greater nuclear charge $(+14)$ than aluminium $(+13)$ .

Hence, the effective nuclear charge exerted on the unpaired 3p electron of silicon would be greater as compared to that of aluminium.

Question

(a) P

The electronic configuration of P: $(1s^2)(2s^2)(2p^6)(3s^2)(3p_{x}^1p_{y}^1p_{z}^1)$ .

Hence, the number of unpaired electrons are 3 in 3p orbital.

Question

(b) Si

The electronic configuration of Si : $(1s^2)(2s^2)(2p^6)(3s^2)(3p_{x}^1p_{y}^1)$ .

Hence, the number of unpaired electrons are 2 in $3p(p_{x}\ and\ p_{y})$ orbital.

Question

(c) Cr

The electronic configuration of Cr : $(1s^2)(2s^2)(2p^6)(3s^2)(3p^6)(3d^5)(4s^1)$ .

Hence, the number of unpaired electrons are 6 (1 in 4s and 5 in 3d) .

Question

(d) Fe

The electronic configuration of Fe : $(1s^2)(2s^2)(2p^6)(3s^2)(3p^6)(3d^6)(4s^2)$ .

Hence, the number of unpaired electrons are 4 (in 3d ) .

Question

(e) Kr

As Krypton (Kr) is a noble gas whose atomic number is 36 and have all orbitals filled.

Hence, there are no unpaired electrons in Kr element.

Question

For a given value of $n$ , $l$ can have values from $0$ to $(n-1)$ .

Therefore, for $n=4$ ,

$l$ can have values from $0$ to $3$ .

i.e., $l = 0,1,2,3$ .

Thus, four subshells are associated with $n=4$ , which are $s,p,d, and \ f.$

The number of orbitals in the $n^{th}$ shell is equal to $n^2$ .

So, for $n=4$ , there are $4^2 =16$ orbitals present.

As each orbital has one electron with spin $m_{s} =-\frac{1}{2}$ .

Hence, there will be $16$ electrons with $m_{s} =-\frac{1}{2}$ .

Topics of NCERT Chapter 2 Chemistry class 11

2.1 Discovery of Sub-atomic Particles

2.2 Atomic Models

2.3 Developments Leading to the Bohr’s Model of Atom

2.4 Bohr’s Model for Hydrogen Atom

2.5 Towards Quantum Mechanical Model of the Atom

2.6 Quantum Mechanical Model of Atom

More About Structure of Atom Class 11 NCERT Chemistry Book Chapter 2

The rich diversity of chemical behavior of various elements can be discovered to the differences in the internal structure of atoms of these elements. This chapter starts with some experimental observations made by scientists towards the end of 19th and beginning of the 20th century. According to these observations, atoms are made of sub-atomic particles, such that protons, electrons and neutrons. This concept is very different from Dalton's atomic theory. NCERT solutions for Class 11 Chemistry chapter 2 Structure of Atom has covered all the topic wise answers as well as the exercise answers.

After completing NCERT solutions for Class 11 Chemistry chapter 2 Structure of Atom, students will be able to know about the discovery of proton, electron and neutron; describe Rutherford, Thomson and Bohr atomic models; understand nature of Planck's quantum theory and electromagnetic radiation; explain the photoelectric effect; state the de Broglie relation and Heisenberg uncertainty principle. Structure of Atom Class 11 Chemistry chapter also explains atomic orbitals in terms of quantum numbers, Pauli exclusion principle, Aufbau principle and Hund's rule of maximum multiplicity and at the end of chapter students will learn how to write the electronic configurations of atoms.

NCERT solutions for class 11 chemistry

 Chapter 1 Some Basic Concepts of Chemistry Chapter-2 Structure of Atom Chapter-3 Chapter-4 Chapter-5 Chapter-6 Chapter-7 Chapter-8 Chapter-9 Chapter-10 Chapter-11 Chapter-12 Chapter-13 Chapter-14

NCERT solutions for class 11 subject wise

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Benefits of NCERT solutions for Class 11 Chemistry Chapter 2 Structure of Atom

• The solutions are written in a comprehensive manner in the NCERT solutions for Class 11 Chemistry chapter 2 Structure of Atom will help you writing answers in your exam.
• Revision will be easy because the detailed solutions will help you to remember the concepts and get you good marks.
• Homework problems will be easier for you, all you need to do is check the detailed NCERT solutions for Class 11 Chemistry chapter 2 Structure of Atom and you are ready to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Also Check NCERT Books and NCERT Syllabus here:

1. Where can I find complete solutions of NCERT Class 11 Chemistry

Refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry  Students can download the chapterwise solutions. Links for chapter wise solutions are provided. For more questions students can use NCERT exemplar.

2. Write is the official website of NCERT

official website of NCERT: http://www.ncert.nic.in/ .

3. What is the weightage of NCERT chapter Structure of Atom for NEET exam

It holds weightage of 2%. NEET is highly compititive exam. So each chapter and topics are important. Solve and analyse previous yearNEET  papers to understand the types of questions asked.

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