Dual Nature Of Radiation And Matter

Chapter 11

DUAL NATURE OF RADIATION AND MATTER

MCQ I

11.1 A particle is dropped from a height $H$. The de Broglie wavelength of the particle as a function of height is proportional to

(a) $H$

(b) $H^{1 / 2}$

(b) $H^{-1 / 2}$

11.2 The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1 MeV$ energy is nearly

(a) $1.2 nm$

(b) $1.2 \times 10^{-3} nm$

(d) $1.2 \times 10^{1} nm$

11.3 Consider a beam of electrons (each electron with energy $E_0$ ) incident on a metal surface kept in an evacuated chamber. Then (a) no electrons will be emitted as only photons can emit electrons.

(b) electrons can be emitted but all with an energy, $E_0$.

(d) electrons can be emitted with any energy, with a maximum of $E_0$.

11.4 Consider Fig. 11.7 in the NCERT text book of physics for Class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of $\theta$ that

(a) will be larger than the earlier value.

(b) will be the same as the earlier value.

(d) will depend on the target.

11.5 A proton, a neutron, an electron and an $\alpha$-particle have same energy. Then their de Broglie wavelengths compare as

(a) $\lambda_p=\lambda_n>\lambda_e>\lambda _{\alpha}$

(b) $\lambda _{\alpha}<\lambda_p=\lambda_n>\lambda_e$

(d) $\lambda_e=\lambda_p=\lambda_n=\lambda _{\alpha}$

11.6 An electron is moving with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ and is in a magnetic field $\mathbf{B}=B_0 \hat{\mathbf{j}}$. Then it’s de Broglie wavelength

(a) remains constant.

(b) increases with time.

(d) increases and decreases periodically.

11.7 An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}(v_0>0)$ is in an electric field $\mathbf{E}=-E_0 \hat{\mathbf{i}}(E_0=.$ constant $.>0)$. It’s de Broglie wavelength at time $t$ is given by

(a) $ \frac{\lambda_0}{(1+\frac{e E_0}{m} \frac{t}{v_0})} $

(b) $\lambda_0(1+\frac{e E_0 t}{m v_0})$

(d) $\lambda_0 t$.

11.8 An electron (mass $m$ ) with an initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}$ is in an electric field $\mathbf{E}=E_0 \hat{\mathbf{j}}$. If $\lambda_0=h / m v_0$, it’s de Breoglie wavelength at time $t$ is given by

(a) $\lambda_0$

(b) $\lambda_0 \sqrt{1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}}}$

(d) $\frac{\lambda_0}{(1+\frac{e^{2} E_0^{2} t^{2}}{m^{2} v_0^{2}})}$

MCQ II

11.9 Relativistic corrections become neccssary when the expression for the kinetic energy $\frac{1}{2} m v^{2}$, becomes comparable with $m c^{2}$, where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

(a) $\lambda=10 nm$

(b) $\lambda=10^{-1} nm$

(d) $\lambda=10^{-6} nm$

11.10 Two particles $A _{1 s}$ and $A_2$ of masses $m_1, m_2(m_1>m_2)$ have the same de Broglie wavelength. Then

(a) their momenta are the same.

(b) their energies are the same.

(d) energy of $A_1$ is more than the energy of $A_2$.

11.11 The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is $v_e=\frac{c}{100}$. Then (a) $\frac{E_e}{E_p}=10^{-4}$

(b) $\frac{E_e}{E_p}=10^{-2}$

(d) $\frac{p_e}{m_e c}=10^{-4}$.

11.12 Photons absorbed in matter are converted to heat. A source emitting $n$ photon/sec of frequency $v$ is used to convert $1 kg$ of ice at $0^{\circ} C$ to water at $0^{\circ} C$. Then, the time $T$ taken for the conversion

(a) decreases with increasing $n$, with $v$ fixed.

(b) decreases with $n$ fixed, $v$ increasing

(d) increases when the product $n v$ increases.

11.13 A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values $\lambda_1$, $\lambda_2$ with $\lambda_1>\lambda_2$. Which of the following statement are true?

(a) The particle could be moving in a circular orbit with origin as centre

(b) The particle could be moving in an elliptic orbit with origin as its focus.

(c) When the de Broglie wave length is $\lambda_1$, the particle is nearer the origin than when its value is $\lambda_2$.

(d) When the de Broglic wavelength is $\lambda_2$, the particle is nearer the origin than when its value is $\lambda_1$.

VSA

11.14 A proton and an $\alpha$-particle are accelerated, using the same potential difference. How are the deBroglie wavelengths $\lambda_p$ and $\lambda_a$ related to each other?

11.15 (i) In the explanation of photo electric effect, we asssume one photon of frequency $v$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E _{\max }$ of the emitted electron as

$ E _{\max }=h v-\phi_0 $

where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $v$ ) what will be the maximum energy for the emitted electron?

(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

11.16 There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.

11.17 Do all the electrons that absorb a photon come out as photoelectrons?

11.18 There are two sources of light, each emitting with a power of $100 W$. One emits X-rays of wavelength $1 nm$ and the other visible light at $500 nm$. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength?

11.19 Consider Fig. 11.1 for photoemission.

How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Fig. 11.1

11.20 Consider a metal exposed to light of wavelength $600 nm$. The maximum energy of the electron doubles when light of wavelength $400 snm$ is used. Find the work function in eV.

11.21 Assuming an electron is confined to a $1 nm$ wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (Ref Eq 11.12 of NCERT Textbook). You can assume the uncertainty in position $\Delta x$ as $1 nm$. Assuming $p \simeq \Delta p$, find the energy of the electron in electron volts.

11.22 Two monochromatic beams A and B of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam $A$ is twice that by beam $B$. Then what inference can you make about their frequencies?

11.23 Two particles A and B of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle $C$. The process conserves momentum. Find the de Broglie wavelength of the particle $C$. (The motion is one dimensional).

11.24 A neutron beam of energy $E$ scatters from atoms on a surface with a spacing $d=0.1 nm$. The first maximum of intensity in the reflected beam occurs at $\theta=30^{\circ}$. What is the kinetic energy $E$ of the beam in $eV$ ?

11.25 Consider a thin target $(10^{-2} m.$ square, $10^{-3} m$ thickness) of sodium, which produces a photocurrent of $100 \mu A$ when a light of intensity $100 W / m^{2}(\lambda=660 nm)$ falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of $Na=0.97 kg / m^{3}$ ].

11.26 Consider an electron in front of metallic surface at a distance $d$ (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\frac{1}{4} \frac{q^{2}}{4 \pi \varepsilon_0 d^{2}}$

Calculate work in taking the charge to an infinite distance from the plate. Taking $d=0.1 nm$, find the work done in electron volts. [Such a force law is not valid for $d<0.1 nm$ ].

11.27 A student performs an experiment on photoelectric effect, using two materials A and B. A plot of $V _{\text{stop }} v s$ $v$ is given in Fig. 11.2.

(i) Which material A or $B$ has a higher work function?

(ii) Given the electric charge of an electron $=1.6 \times 10^{-19} C$, find the value of $h$ obtained from the experiment for both $A$ and $B$.

Comment on whether it is consistent with Einstein’s theory:

Fig. 11.2

11.28 A particle A with a mass $m_A$ is moving with a velocity $v$ and hits a particle B (mass $m_B$ ) at rest (one dimensional motion). Find the change in the de Broglic wavelength of the particle A. Treat the collision as elastic.

11.29 Consider a $20 W$ bulb emitting light of wavelength $5000 \AA$ and shining on a metal surface kept at a distance $2 m$. Assume that the metal surface has work function of $2 eV$ and that each atom on the metal surface can be treated as a circular disk of radius $1.5 A$.

(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]

(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomc disk to receive energy equal to work function ( $2 eV$ )?

(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?

(v) Can you explain how photoelectric effect was observed instantaneously?

[Hint: Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say $1 cm^{2}$ and estimate what would happen?]

Chapter 11

11.1 (d)

11.2 (b)

11.3 (d)

11.4 (c)

11.5 (b)

11.6 (a)

11.7 (a)

11.8 (c)

11.9 (c), (d)

11.10 (a), (c)

11.11 (b), (c)

11.12 (a), (b), (c)

11.13 (b), (d)

11.14 $ \quad \lambda_p / \lambda_d=p_x / p_p=\frac{\sqrt{2 m _{\alpha} E _{\alpha}}}{\sqrt{2 m_p E_p}}=\sqrt{8}: 1$

11.15 (i) $E _{\max }=2 h v-\phi$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence such emissions will be negligible.

11.16 In the first case energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances.

11.17 No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

11.18 Total $E$ is constant

Let $n_1$ and $n_2$ be the number of photons of X-rays and visible region

$n_1 E_1=n_2 E_2$

$n_1 \frac{h c}{\lambda_1}=n_2 \frac{h c}{\lambda_2}$

$\frac{n_1}{n_2}=\frac{\lambda_1}{\lambda_2}$

$\frac{n_1}{n_2}=\frac{1}{500}$.

11.19 The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons.

11.20 Maximum energy $=h v-\phi$

$(\frac{1230}{600}-\phi)=\frac{1}{2}(\frac{1230}{400}-\phi)$

$\phi=\frac{1230}{1200}=1.02 eV$.

11.21 $ \Delta x \Delta p \simeq \hbar$

$\Delta p \simeq \frac{\hbar}{\Delta x} \simeq \frac{1.05 \times 10^{-34} Js}{10^{-9} m}=1.05 \times 10^{-25}$

$E=\frac{p^{2}}{2 m}=\frac{(1.05 \times 10^{-25})^{2}}{2 \times 9.1 \times 10^{-31}}=\frac{1.05^{2}}{18.2} \times 10^{-19} J=\frac{1.05^{2}}{18.2 \times 1.6} eV$

$=3.8 \times 10^{-2} eV$

11.22 $ I=n_A n_A=n_B v_B$

$\frac{n_A}{n_B}=2=\frac{v_B}{v_A}$

The frequency of beam $B$ is twice that of $A$.

11.23

$p_c=|p_A|+|p_B|=\frac{h}{\lambda_A}+\frac{h}{\lambda_B}=\frac{h}{\lambda_c}=\frac{h}{\lambda_c}$ if $p_A, p_B>0$ or $p_A, p_B<0$

or $\lambda_c=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$

If $p_A>0, p_B<0$ or $p_A<0, p_B>0$

$p_c=h \frac{\lambda_B-\lambda_A}{|\lambda_A \cdot \lambda_B|}=\frac{h}{\lambda_c}$

$\lambda_c=\frac{\lambda_B \cdot \lambda_A}{|\lambda_A-\lambda_B|}$.

11.24 $2 d \sin \theta=\lambda=d=10^{-10} m$.

$ \begin{gathered} p=\frac{h}{10^{-10}}=\frac{6.6 \times 10^{-34}}{10^{-10}}=6.6 \times 10^{-21} kg m / s \\ E=\frac{(6.6 \times 10^{-24})^{2}}{2 \times(1.7 \times 10^{-27})} \times 1.6 \times 10^{-19}=\frac{6.6^{2}}{2 \times 1.7} \times 1.6 \times 10^{-2} eV \\ =20.5 \times 10^{-2} eV=0.21 eV \end{gathered} $

11.25 $6 \times 10^{26} Na$ atoms weighs $23 kg$.

Volume of target $=(10^{-4} \times 10^{-3})=10^{-7} m^{3}$

Density of sodium $=(d)=0.97 kg / m^{3}$

Volume of $6 \times 10^{26} Na$ atoms $=\frac{23}{0.97} m^{3}=23.7 m^{3}$

Volume occupied of $1 Na$ atom $=\frac{23}{0.97 \times 6 \times 10^{26}} m^{3}=3.95 \times 10^{-26} m^{3}$

No. of sodium atoms in the target $=\frac{10^{-7}}{3.95 \times 10^{-26}}=2.53 \times 10^{18}$

Number of photons/s in the beam for $10^{-4} m^{2}=n$

Energy per s $n h v=10^{-4} J \times 100=10^{-2} W$

$h v($ for $\lambda=660 nm)=\frac{1234.5}{600}$

$ =2.05 eV=2.05 \times 1.6 \times 10^{-19}=3.28 \times 10^{-19} J $

$n=\frac{10^{-2}}{3.28 \times 10^{-19}}=3.05 \times 10^{16} / s$

$n=\frac{1}{3.2} \times 10^{17}=3.1 \times 10^{16}$

If $P$ is the probability of emission per atom, per photon, the number of photoelectrons emitted/second

$=P \times 3.1 \times 10^{16} \times 2.53 \times 10^{18}$

Current $=P \times 3.1 \times 10^{+16} \times 2.53 \times 10^{18} \times 1.6 \times 10^{-19} A$

$ =P \times 1.25 \times 10^{+16} A $

This must equal $100 \mu A$ or

$P=\frac{100 \times 10^{-6}}{1.25 \times 10^{+16}}$

$\therefore P=8 \times 10^{-21}$

Thus the probability of photemission by a single photon on a single atom is very much less than 1 . (That is why absorption of two photons by an atom is negligible).

11.26 Work done by an external agency $=+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{1}{4} \int_d^{\infty} \frac{q^{2}}{x^{2}} d x=\frac{1}{4} \cdot \frac{q^{2}}{4 \pi \varepsilon_0 d}$

With $d=0.1 nm$, energy $=\frac{(1.6 \times 10^{-19}) \times 9 \times 10^{9}}{4(10^{-10}) \times 1.6 \times 10^{-19}} eV$

$ =\frac{1.6 \times 9}{4} eV=3.6 eV $

11.27 (i) Stopping potential $=0$ at a higher frequency for B. Hence it has a higher work function.

(ii) Slope $=\frac{h}{e}=\frac{2}{(10-5) \times 10^{14}}$ for A.

$ =\frac{2.5}{(15-10) \times 10^{14}} \text{ for } B \text{. } $

$ \begin{aligned} h= & \frac{1.6 \times 10^{-19}}{5} \times 2 \times 10^{-14}=6.04 \times 10^{-34} Js \text{ for } A \\ & =\frac{1.6 \times 10^{-19} \times 2.5 \times 10^{-14}}{5}=8 \times 10^{-34} Js \text{ for } B . \end{aligned} $

Since $h$ works out differently, experiment is not consistent with the theory.

11.28

$m_A v=m_A v_1+m_B v_2$

$\frac{1}{2} m_A v^{2}=\frac{1}{2} m_A v_1^{2}+\frac{1}{2} m_B v_2^{2}$

$\therefore \frac{1}{2} m_A(v-v_1)(v_A+v_1)=\frac{1}{2} m_B v_B^{2}$

$\therefore v+v_1=v_2$

or $v=v_2-v_1$

$\therefore v_1=(\frac{m_A-m_B}{m_A+m_B}) v, \quad$ and $\quad v_2=(\frac{2 m_A}{m_A+m_B}) v$

$\therefore \lambda _{\text{initial }}=\frac{h}{m_A v}$

$\lambda _{\text{final }}=\frac{h}{m_A v}=|\frac{h(m_A+m_B)}{m_A(m_A-m_B) v}|$

$\therefore \Delta \lambda=\frac{h}{m_A v}[|\frac{.m_A+m_B)}{m_A-m_B}|-1]$

11.29 (i) $\frac{d N}{d t}=\frac{P}{(h c / \lambda)}=5 \times 10^{19} / s$

(ii) $\frac{h c}{\lambda}=2.49 eV>W_0:$ Yes.

(iii) P. $\frac{\pi r^{2}}{4 \pi d^{2}} \Delta t=W_0, \Delta t=28.4 s$

(iv) $N=(\frac{d N}{d t}) \times \frac{\pi r^{2}}{4 \pi d^{2}} \times \Delta t=2$