NCERT Solutions for Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter

Class 12 Physics Chapter 11 - Dual nature of radiation and matter: Additional Questions

The Class 12 Physics Chapter 11 - Dual nature of radiation and matter is an introduction to one of the most interesting findings of modern physics - the dual nature of light and matter, i.e. they can act like waves and like particles at the same time. This theory fills in the gap between quantum and classical physics. The extra problems in this chapter enable the students to reinforce their knowledge on such concepts as the photoelectric effect, de Broglie wavelength and photoelectric equation as proposed by Einstein to ensure an in-depth comprehension on the major concepts of the exams and their real-life use.

Q1.a) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$. Find their momenta.

Answer:

Their momenta depend only on the de Broglie wavelength; therefore, it will be the same for both the electron and the photon

$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{10^{-9}} \\
p & =6.62 \times 10^{-25} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

b) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$. Find the energy of the photon.

Answer:

The energy of the photon is given by

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

h is the Planks constant, c is the speed of light, and lambda is the wavelength

$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-9}}\\$

$E=1.86\times 10^{-16}\ J$

c) An electron and a photon each have a wavelength of $1.00\hspace{1mm}nm$. Find the kinetic energy of an electron.

Answer:

The kinetic energy of the electron is. In the equation below, p is the momentum

$K=\frac{p^{2}}{2m_{e}}$
$K=\frac{(6.62\times 10^{-25})^{2}}{2\times 9.1\times 10^{-31}}$
$K=2.41\times 10^{-19}\ J$

Q2.a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.4*10-10 m?

Answer:

For the given wavelength momentum of the neutron will be $p$ given by
$\begin{aligned}
p & =\frac{h}{\lambda} \\
p & =\frac{6.62 \times 10^{-34}}{1.4 \times 10^{-10}} \\
p & =4.728 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

The kinetic energy K would therefore be
$\begin{aligned}
K & =\frac{p^2}{2 m} \\
K & =\frac{\left(4.728 \times 10^{-24}\right)^2}{2 \times 1.675 \times 10^{-27}} \\
K & =6.67 \times 10^{-21} J
\end{aligned}$

b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of $(3/2)\hspace{1mm}kT$ at $300\hspace{1mm}K$.

Answer:

The kinetic energy of the neutron is
$\begin{aligned}
& K=\frac{3}{2} k T \\
& K=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \\
& K=6.21 \times 10^{-21} J
\end{aligned}$

Where k Boltzmann's Constant is $1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$
The momentum of the neutron will be $p$
$\begin{aligned}
& p=\sqrt{2 m_N K} \\
& p=\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \\
& p=4.56 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

The associated De Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{4.56 \times 10^{-24}} \\
\lambda & =1.45 \times 10^{-10} \mathrm{~m}
\end{aligned}$

The de Broglie wavelength of the neutron is 0.145 nm.

Q3. What is the de Broglie wavelength of a nitrogen molecule in air at $300\hspace{1mm}K$ ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $=14.0076\hspace{1mm} \mu$ )

Answer:

Since the molecule is moving with the root-mean-square speed, the kinetic energy $K$ will be given by $\mathrm{K}=3 / 2 \mathrm{kT}$ where k is the Boltzmann's constant and T is the absolute Temperature

In the given case Kinetic Energy of a Nitrogen molecule will be
$\begin{aligned}
& K=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \\
& K=6.21 \times 10^{-21} J
\end{aligned}$

$\text { Mass of Nitrogen molecule }=2 \times 14.0076 \times 1.66 \times 10^{-27}=4.65 \times 10^{-26} \mathrm{~kg}$

The momentum of the molecule is
$\begin{aligned}
& p=\sqrt{2 m K} \\
& p=\sqrt{2 \times 4.65 \times 10^{-26} \times 6.21 \times 10^{-21}} \\
& p=2.4 \times 10^{-23} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$

The associated de Broglie wavelength is
$\begin{aligned}
\lambda & =\frac{h}{p} \\
\lambda & =\frac{6.62 \times 10^{-34}}{2.4 \times 10^{-23}} \\
\lambda & =2.75 \times 10^{-11} \mathrm{~m}
\end{aligned}$

The nitrogen molecule will have a De Broglie wavelength of 0.0275 nm.

Q4.a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\hspace{1mm}V$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its $e/m$ is given to be $1.76 \times 10^{11} \mathrm{~kg}^{-1}$

Answer:

The kinetic energy of an electron accelerated through a Potential Difference V is K=eV where e is the electronic charge.

The speed of the electrons after being accelerated through a potential difference of 500 V will be

$v=\sqrt{\frac{2K}{m_{e}}}$
$v=\sqrt{\frac{2eV}{m_{e}}}$
$v=\sqrt{2\times 1.76\times 10^{11}\times 500}$
$ v=1.366\times 10^{7}ms^{-1}$
Specific charge is e/$m_e$ =1.366 $\times$ $10^{11}$ C/kg

Q4.b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Answer:

Using the same formula, we get the speed of electrons to be 1.88 $\times$ $10^{9}$ m/s. This is wrong because the speed of the electron is coming out to be more than the speed of light. This discrepancy is occurring because the electron will be travelling at a very large speed, and in such cases(relativistic), the mass of the object cannot be taken to be the same as the rest mass.

In such a case

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m 0 is the rest mass of the body, v is the very high speed at which the body is travelling, and c is the speed of light.

Q5.a) A monoenergetic electron beam with an electron speed of $5.20 \times 10^6 \mathrm{~ms}^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} T$ normal to the beam velocity. What is the radius of the circle traced by the beam, given $e/m$ for electron equals $1.76 \times 10^{11} \mathrm{Ckg}^{-1}$

Answer:

The force due to the magnetic field on the electron will be $F_b=e v B$ (since the angle between the velocity and magnetic field is $90^{\circ}$ )

This $\mathrm{F}_{\mathrm{b}}$ acts as the centripetal force required for circular motion. Therefore
$\begin{aligned}
& F_b=\frac{m v^2}{r} \\
& e v B=\frac{m v^2}{r} \\
& r=\frac{m v}{e B} \\
& r=\frac{5.2 \times 10^6}{1.76 \times 10^{11} \times 1.3 \times 10^{-4}} \\
& r=0.227 m
\end{aligned}$

Q5.b) Is the formula you employ in (a) valid for calculating the radius of the path of a $20\hspace{1mm}MeV$ electron beam? If not, in what way is it modified?

Answer:

The formula used in (a) can not be used. As the electron would be travelling at a very high speed, we can not take its mass to be equal to its rest mass, as its motion won't be within the non-relativistic limits.

The value for the mass of the electron would get modified to

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m 0 is the rest mass of the body, v is the very high speed at which the body is travelling, and c is the speed of light.

The radius of the circular path would be

$r=\frac{m_{e}v}{eB\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Q6. An electron gun with its collector at a potential of $100\hspace{1mm}V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( $10^{-2} \mathrm{~mm}$ of Hg). A magnetic field of $2.83\times 10^4\hspace{1mm}T$ curves the path of the electrons in a circular orbit of radius $12.0\hspace{1mm}cm$ (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) . Determine $e/m$ from the data.

Answer:

The kinetic energy of an electron after being accelerated through a potential difference of V volts is eV, where e is the electronic charge.

The speed of the electron will become

$v=\sqrt{\frac{2eV}{m_{e}}}$

Since the magnetic field curves, the path of the electron in a circular orbit, the electron's velocity must be perpendicular to the magnetic field.

The force due to the magnetic field is therefore F =evB

This magnetic force acts as a centripetal force. Therefore

$\begin{aligned}
& \frac{m_e v^2}{r}=e v B \\
& \frac{m_e v}{r}=e B \\
& \frac{m_e}{r} \times \sqrt{\frac{2 e V}{m_e}}=e B \\
& \sqrt{\frac{e}{m_e}}=\frac{\sqrt{2 V}}{B r} \\
& \frac{e}{m_e}=\frac{2 V}{r^2 B^2} \\
& \frac{e}{m_e}=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^2 \times(0.12)^2} \\
& \frac{e}{m_e}=1.73 \times 10^{11} C~kg^{-1}
\end{aligned}$

Q7.a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at $0.45\dot{A}$. What is the maximum energy of a photon in the radiation?

Answer:

The wavelength of photons with maximum energy=0.45 $A^{\circ}$

The energy of the photons is

$\begin{aligned} & E=\frac{h c}{\lambda} \\ & E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10}} \\ & E=4.413 \times 10^{-15} \mathrm{~J} \\ & E=27.6 \mathrm{keV}\end{aligned}$

b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

Answer:

In such a tube where an X-ray of energy 27.6 keV is to be produced, the electrons should have an energy about the same value, and therefore accelerating voltage should be of the order 30 keV.

Q8. In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy $10.2\hspace{1mm}BeV$ into two $\gamma$ -rays of equal energy. What is the wavelength associated with each $\gamma$ -ray? ( $1\hspace{1mm}BeV=10^9\hspace{1mm}eV$ )

Answer:

The total energy of 2 $\gamma$ rays=10.2 BeV

The average energy of 1 $\gamma$ ray, E=5.1 BeV

The wavelength of the gamma-ray is given by

$\begin{aligned} & \lambda=\frac{c}{\nu} \\ & \lambda=\frac{h c}{h \nu} \\ & \lambda=\frac{h c}{E} \\ & \lambda=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5.1 \times 10^9 \times 1.6 \times 10^{-19}} \\ & \lambda=2.436 \times 10^{-16} \mathrm{~m}\end{aligned}$

Q9.a) Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. The number of photons emitted per second by a Medium wave transmitter of $10\hspace{1mm}kW$ power, emitting radiowaves of wavelength $500\hspace{1mm}m$ .

Answer:

The power emitted by the transmitter $(P)=10 \mathrm{~kW}$

Wavelengths of photons being emiited $=500 \mathrm{~m}$

The energy of one photon is $E$

$\begin{aligned}
E & =\frac{h c}{\lambda} \\
E & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{500} \\
E & =3.96 \times 10^{-28} \mathrm{~J}
\end{aligned}$

The number of photons emitted per second(n) is given by
$\begin{aligned}
n & =\frac{P}{E} \\
n & =\frac{10000}{3.96 \times 10^{-28}} \\
n & =2.525 \times 10^{31} \mathrm{~s}^{-1}
\end{aligned}$

Q9.b) Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $\left(\sim 10^{-10} \mathrm{Wm}^2\right)$. Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about $6 \times 10^{14} \mathrm{~Hz}$.

Answer:

The minimum perceivable intensity of white light(I)=$10^{-10}$ $Wm^{-2}$

Area of the pupil(A)=0.4 cm 2 =4 $\times$ $10^{-5}$ $m^{2}$

Power of light falling on our eyes at the minimum perceivable intensity is P

P=IA

P=$10^{-10}$ $\times$ 4 $\times$ $10^{-5}$

P=4 $\times$ $10^{-15}$ W

The average frequency of white light( $\nu$ )=6 $\times$ $10^{14}$ Hz

The average energy of a photon in white light is

$\\E=h\nu$
$E=6.62\times 10^{-34}\times 6\times 10^{14}$
$E=3.972\times 10^{-19} J$

The number of photons reaching our eyes is n

$n=\frac{P}{E}$
$n=\frac{4\times 10^{-15}}{3.972\times 10^{-19}}$
$n=1.008\times 10^{4}s^{-1}$

Q10. Ultraviolet light of wavelength $2271\hspace{1mm}\dot{A}$ from a $100\hspace{1mm}W$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3\hspace{1mm}V$, estimate the work function of the metal. How would the photo-cell respond to a high intensity $(\sim 10^5\hspace{1mm}Wm^2)$ red light of wavelength $6382\hspace{1mm}\dot{A}$ produced by a $He-Ne$ laser?

Answer:

The energy of the incident photons is E given by

$E=\frac{hc}{\lambda }$
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{2271\times 10^{-10}\times 1.6\times 10^{-19}}$
$E=5.465 eV$

Since the stopping potential is -1.3 V work function is

$\phi _{0}=5.465-1.3$
$\phi _{0}=4.165 eV$

The energy of photons, which red light consists of, is E_R

$E_{R}=\frac{hc}{\lambda _{R}}$
$E_{R}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{6382\times 10^{-10}\times 1.6\times 10^{-19}}$
$ E_{R}=1.945eV$

Since the energy of the photons which red light consists of has less energy than the work function, there will be no photoelectric emission when they are incident.

Q11. Monochromatic radiation of wavelength $640.2 \mathrm{~nm}\left(1\mathrm{~nm}=10^{-9} \mathrm{~m}\right)$from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be $0.54\hspace{1mm}V$ . The source is replaced by an iron source and its $427.2\hspace{1mm}nm$ line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

The wavelength of photons emitted by the neon lamp $=640.2 \mathrm{~nm}$

The energy of photons emitted by the neon lamp is E given by
$\begin{aligned}
& E_1=\frac{h c}{\lambda} \\
& E_1=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \\
& E_1=1.939 \mathrm{eV}
\end{aligned}$

Stopping potential is 0.54 V
Work function is therefore
$\begin{aligned}
\phi_0 & =1.939-0.54 \\
\phi_0 & =1.399 \mathrm{eV}
\end{aligned}$

The wavelength of photons emitted by the iron source $=427.2 \mathrm{~nm}$
The energy of photons emitted by the ion source is
$\begin{aligned}
& E_2=\frac{h c}{\lambda} \\
& E_2=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \\
& E_2=2.905 \mathrm{eV}
\end{aligned}$

The new stopping voltage is
$E_2-\phi_0=2.905-1.399=1.506 \mathrm{~V}$

Q12. A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives several spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

$\lambda_1=3650 \dot{A}, \lambda_2=4047 \dot{A}, \lambda_3=4358 \dot{A}, \lambda_4=5461 \dot{A}, \lambda_5=6907 \dot{A}$

The stopping voltages, respectively, were measured to be

$V_{01}=1.28 \mathrm{~V}, V_{02}=0.95 \mathrm{~V}, V_{03}=0.74 \mathrm{~V}, V_{04}=0.16 \mathrm{~V}, V_{05}=0 \mathrm{~V}$

Determine the value of Planck's constant $h$, the threshold frequency and the work function for the material.

Answer:

$h\nu =\phi _{0}+eV$
$V=(\frac{h}{e})\nu -\phi_{0}\\$

where V is the stopping potential, h is Planck's constant, e is the electronic charge, $\nu$ is the frequency of incident photons and $\phi _{0}$ is the work function of the metal in electron Volts.

To calculate Planck's constant from the above data, we plot the stopping potential vs frequency graph

$\nu_{1}=\frac{c}{\lambda_{1} }=\frac{3\times 10^{8}}{3650\times 10^{-10}}=8.219\times 10^{14}\ Hz$

$\nu_{2}=\frac{c}{\lambda_{2} }=\frac{3\times 10^{8}}{4047\times 10^{-10}}=7.412\times 10^{14}\ Hz$

$\nu_{3}=\frac{c}{\lambda_{3} }=\frac{3\times 10^{8}}{4358\times 10^{-10}}=6.884\times 10^{14}\ Hz$

$\nu_{4}=\frac{c}{\lambda_{4} }=\frac{3\times 10^{8}}{5461\times 10^{-10}}=5.493\times 10^{14}\ Hz$

$\nu_{5}=\frac{c}{\lambda_{5} }=\frac{3\times 10^{8}}{6907\times 10^{-10}}=4.343\times 10^{14}\ Hz$

The plot we get is

From the above figure, we can see that the curve is almost a straight line.

The slope of the above graph will give Planck's constant divided by the electronic charge. The Planck's constant calculated from the above chart is

$h=\frac{\left ( 1.28-0.16 \right )\times 1.6\times 10^{-19}}{(8.214-5.493)\times 10^{14}}$
$h=6.573\times 10^{-34} Js$

Planck's constant calculated from the above chart is therefore $6.573\times 10^{-34}\ Js$

Q13. The work function for the following metals is given: $Na:2.75\hspace{1mm}eV, K:2.30\hspace{1mm}eV, Mo:4.17\hspace{1mm}eV ,Ni:5.15\hspace{1mm}eV.$

Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\hspace{1mm}\dot{A}$ from a $He-Cd$ laser placed $1\hspace{1mm}m$ away from the photocell? What happens if the laser is brought nearer and placed $50\hspace{1mm}cm$ away?

Answer:

The wavelength of the incident photons= $3300\dot{A}$

The energy of the incident photons is

$E=\frac{hc}{\lambda }$
$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3300\times 10^{-10}\times 1.6\times 10^{-19}}$
$ E=3.16 eV$

Mo and Ni will not give photoelectric emission for radiation of wavelength $3300\hspace{1mm}\dot{A}$ from a $He-Cd$.

If the laser is brought nearer, no change will be there in case of Mo and Ni, although there will be more photoelectrons in case of Na and K.

Q14. Light of intensity $10^{-5} \mathrm{Wm}^{-2}$ falls on a sodium photo-cell of surface area $2\hspace{1mm}cm^2$ . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2\hspace{1mm}eV$ . What is the implication of your answer? (Effective atomic area of a sodium atom = 10 -20 m 2 )

Answer:

Intensity of Incident light(I) = $10^{-5} \mathrm{Wm}^{-2}$

The surface area of the sodium photocell (A)=2 cm 2 = 2 $\times$ $10^{-4}$ $m^{2}$

The rate at which energy falls on the photo cell=IA=2 $\times$ $10^{-9}$ W

The rate at which each of the 5 surfaces absorbs energy= IA/5=4 $\times$ $10^{-10}$ W

Effective atomic area of a sodium atom (A')= $10^{-20}$ $m^{2}$

The rate at which each sodium atom absorbs energy is given by R

$\begin{aligned} R & =\frac{I A}{5} \times \frac{A^{\prime}}{A} \\ R & =\frac{10^{-5} \times 10^{-20}}{5} \\ R & =2 \times 10^{-26} \mathrm{~J} / \mathrm{s}\end{aligned}$

The time required for photoelectric emission is

$\begin{aligned} t & =\frac{\phi_0}{R} \\ t & =\frac{2 \times 1.6 \times 10^{-19}}{2 \times 10^{-26}} \\ t & =1.6 \times 10^7 \mathrm{~s} \\ t & \approx 0.507 \text { years }\end{aligned}$

Q15. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to $1 \dot{A}$, which is of the order of inter-atomic spacing in the lattice) $\left(m_e=9.11 \times 10^{-31} \mathrm{~kg}\right)$.

Answer:

According to De Broglie's equation

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

The kinetic energy of an electron with De Broglie wavelength $1\hspace{1mm}\dot{A}$ is given by

$\begin{aligned} K & =\frac{p^2}{2 m_e} \\ K & =\frac{h^2}{\lambda^2 2 m_e} \\ K & =\frac{\left(6.62 \times 10^{-34}\right)^2}{2 \times 10^{-20} \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19}} \\ K & =149.375 \mathrm{eV}\end{aligned}$

The kinetic energy of photon having wavelength $1\hspace{1mm}\dot{A}$ is

$\begin{aligned} & E=\frac{h c}{\lambda} \\ & E=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \\ & E=12.375 \mathrm{keV}\end{aligned}$

Therefore, for the given wavelength, a photon has much higher energy than an electron.

Q16.a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150\hspace{1mm}eV$ . As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.$\left(m_e=9.11 \times 10^{-31} \mathrm{~kg}\right)$

Answer:

Kinetic energy of the neutron(K)=150eV

The de Broglie wavelength associated with the neutron is

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_N K}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 150 \times 1.6 \times 10^{-19}}} \\ & \lambda=2.327 \times 10^{-12} \mathrm{~m}\end{aligned}$

Since an electron beam with the same energy has a wavelength much larger than the above-calculated wavelength of the neutron, a neutron beam of this energy is not suitable for crystal diffraction, as the wavelength of the neutron is not of the order of the dimension of interatomic spacing.

b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature $(27\hspace{1mm}^{\circ}C )$. Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Answer:

Absolute temperature = 273+27=300K

Boltzmann's Constant=1.38 $\times$ $10^{-23}$ J/mol/K

The de Broglie wavelength associated with the neutron is

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_N K}} \\ & \lambda=\frac{h}{\sqrt{3 k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=1.446 \dot{A}\end{aligned}$

Since this wavelength is comparable to the order of interatomic spacing of a crystal, it can be used for diffraction experiments. The neutron beam is to be thermalised so that its de Broglie wavelength attains a value such that it becomes suitable for the crystal diffraction experiments.

Q17. An electron microscope uses electrons accelerated by a voltage of $50\hspace{1mm}kV$. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Answer:

The potential difference through which electrons are accelerated(V)=50kV.

Kinetic energy(K) of the electrons would be eV, where e is the electronic charge

The De Broglie wavelength associated with the electrons is

$\begin{aligned} & \lambda=\frac{h}{\sqrt{2 m_e K}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50000}} \\ & \lambda=5.467 \times 10^{-12} \mathrm{~m}\end{aligned}$

The wavelength of yellow light = 5.9 $\times$ $10^{-7}$ m

The calculated De Broglie wavelength of the electron microscope is about 10⁵ times more than that of yellow light and since resolving power is inversely proportional to the wavelength, the resolving power of the electron microscope is roughly 10⁵ times that of an optical microscope.

Q18. The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams.

(Rest mass energy of electron $=0.511\hspace{1mm}MeV$ .)

Answer:

Rest mass of the electron

$=mc^2=0.511MeV$

Momentum

$P=\frac{h}{\lambda}=\frac{6.63\times 10^{-34}}{10^{-15}}$

using the relativistic formula for energy

$E^2=(CP)^2+(mc^2)^2$

$=(3\times10^8 \times 6.63\times 10^{-19})^2+(0.511\times1.6\times10^{-19})^2$

$\approx 1.98\times10^{-10} J$

Q19. Find the typical de Broglie wavelength associated with a $He$ atom in helium gas at room temperature ( $27\hspace{1mm}^{\circ}C$ ) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Answer:

The kinetic energy K of a He atom is given by

$K=\frac{3}{2}kT$

m He, i.e. mass of one atom of H,e can be calculated as follows

$m_{He}=\frac{4\times 10^{-3}}{N_{A}} =\frac{4\times 10^{-3}}{6.023\times 10^{23}}=6.64\times 10^{-27} kg$ (N A is the Avogadro's Number)

The de Broglie wavelength is given by

$\begin{aligned} & \lambda=\frac{h}{p} \\ & \lambda=\frac{h}{\sqrt{2 m_{H e} K}} \\ & \lambda=\frac{h}{\sqrt{3 m_{H e} k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=7.27 \times 10^{-11} \mathrm{~m}\end{aligned}$

The mean separation between two atoms is given by the relation

$d=\left ( \frac{V}{N} \right )^{\frac{1}{3}}\\$

From the ideal gas equation, we have

$\begin{aligned} & P V=n R T \\ & P V=\frac{N R T}{N_A} \\ & \frac{V}{N}=\frac{R T}{P N_A}\end{aligned}$

The mean separation is therefore

$\begin{aligned} & d=\left(\frac{R T}{P N_A}\right)^{\frac{1}{3}} \\ & d=\left(\frac{k T}{P}\right)^{\frac{1}{3}} \\ & d=\left(\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^5}\right)^{\frac{1}{3}} \\ & d=3.35 \times 10^{-9} \mathrm{~m}\end{aligned}$

The mean separation is greater than the de Broglie wavelength.

Q20. Compute the typical de Broglie wavelength of an electron in a metal at $27\hspace{1mm}^{\circ}C$ and compare it with the mean separation between two electrons in a metal which is given to be about 2*10-10m.

Answer:

The de Broglie wavelength associated with the electrons is

$\begin{aligned} & \lambda=\frac{h}{\sqrt{3 m_e k T}} \\ & \lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3 \times 9.1 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=6.2 \times 10^{-9} \mathrm{~nm}\end{aligned}$

The de Broglie wavelength of the electrons is comparable to the mean separation between two electrons.

Answer the following questions:

Q21.a) Quarks inside protons and neutrons are thought to carry fractional charges $[(+2/3)e;(-1/3)e]$ . Why do they not show up in Millikan’s oil-drop experiment?

Answer:

Quarks are thought to be tightly bound within a proton or neutron by forces which grow stronger if one tries to pull them apart. That is even though fractional charges may exist in nature, the observable charges are still integral multiples of the charge of the electron

Answer the following questions:

b) What is so special about the combination $e/m$? Why do we not simply talk of $e$ and $m$ separately?

Answer:

The speed of a charged particle is given by the relations

$v=\sqrt{2K\left ( \frac{e}{m} \right )}$

or

$v=Br\left ( \frac{e}{m} \right )$

As we can see, the speed depends on the ratio e/m, which is of such huge importance.

Answer the following questions:

c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

Answer:

At ordinary pressure, due to a large number of collisions among themselves, the gases have no chance of reaching the electrodes, while at very low pressure, these collisions decrease exponentially and the gas molecules have a chance of reaching the respective electrodes and therefore are capable of conducting electricity.

d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if the incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

Answer:

The work function is defined as the minimum energy below which an electron will never be ejected from the metal. But when photons with high energy are incident, it is possible that electrons from different orbits get ejected and would, therefore, come out of the atom with different kinetic energies.

e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
$E=hv,p=\frac{h}{\lambda }$
But while the value of $\lambda$ is physically significant, the value of $v$ (and therefore, the value of the phase speed $v \lambda$ ) has no physical significance. Why?

Answer:

The absolute energy has no significance because of the reference point being arbitrary, and thus the inclusion of an arbitrary constant renders the value of $\nu\lambda$ and $\nu$ to have no physical significance as such.

The group speed is defined as

$V_{G}=\frac{h}{\lambda m}$

Due to the significance of the group speed, the absolute value of the wavelength has physical significance.

Class 12 Physics Chapter 11 - Dual nature of radiation and matter: Higher Order Thinking Skills (HOTS) Questions

The Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter HOTS (Higher Order Thinking Skills) questions are designed to challenge students to think beyond basic definitions and formulas. These questions test conceptual clarity on topics like photoelectric effect, wave-particle duality, and electron emission, encouraging analytical and application-based learning. Solving these HOTS questions helps students develop strong problem-solving skills and boosts their confidence for competitive exams like JEE and NEET.

Q.1 Light of wavelength 6200 Å falls on a metal having a photoelectric work function of 2 eV. What is the value of stopping potential?

Answer:

Energy corresponding to $6200 Å=\frac{12375}{6200} \mathrm{ev}=1.996 \mathrm{eV}=2 \mathrm{eV}$
The electron is emitted with 2 eV kinetic energy. So, the stopping potential is 2V.
So, the stopping potential is 2V.

Q.2 The photoelectric threshold frequency of a metal is $\nu$. When light of frequency $4 \nu$ is incident on the metal, the maximum kinetic energy of the emitted photoelectrons is -

Answer:

$\begin{aligned} & \mathrm{E}=\phi+\mathrm{K}_{\max } \\ & \mathrm{K}_{\max }=4 \mathrm{hv}-\mathrm{h} v=3 \mathrm{~h} v\end{aligned}$

Q.3 The de-Broglie wavelength of a proton $\left(\right.$ mass $\left.=1.6 \times 10^{-27} \mathrm{~kg}\right)$ accelerated through a potential difference of 1 kV is
Answer:
$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}} \\ & \therefore \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times\left(1.6 \times 10^{-27}\right) \times\left(1.6 \times 10^{-19}\right) \times 1000}} \\ & \text { or } \lambda=\frac{6.6 \times 10^{-34} \times 10^{22}}{\sqrt{1.6 \sqrt{20}}}=0.9 \times 10^{-12} \mathrm{~m}\end{aligned}$

Q.4 The de-Broglie wavelength of a neutron at $927^{\circ} \mathrm{C}$ is $\lambda$. What will be its wavelength at $27^{\circ} \mathrm{C}$?
Answer:

The de-Broglie wavelength of a material particle at temperature $T$ is given by

$
\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mkT}}}, \text { where is Boltzmann's constant. } \\
& \Rightarrow \lambda \propto \frac{1}{\sqrt{T}} \\
& \therefore \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
& \text { or } \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{1200}{300}}=2 \\
& \therefore \lambda_2=2 \lambda_1=2 \lambda
\end{aligned}
$

Q.5 What is the de Broglie wavelength of the wave associated with an electron that has been accelerated through a potential difference of 50V?
Answer:

de - Broglie wavelength $(\lambda)$ is independent of charge
The gain of kinetic energy by an electron is eV

$
\begin{aligned}
& \frac{1}{2} m v^2=e V \\
& v=\sqrt{\frac{2 e V}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 50}{9.11 \times 10^{-31}}} \\
& =4.19 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& \therefore \lambda=\frac{h}{m v}=\frac{6.62 \times 10^{-34}}{9.1 \times 10^{-31} \times 4,19 \times 10^6} \\
& =1.74 \times 10^{-10} \mathrm{~m}
\end{aligned}
$