Wave Optics JEE Advanced PYQ | Chapter-wise Physics Previous Year Questions

Q 1 :

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? $[Take \frac{h c}{e} = 1.24 \times 10^{- 6} J m C^{- 1}]$ [2022]

$1.72 \times 10^{- 7} m, 1.20 eV$
$1.72 \times 10^{- 7} m, 5.60 eV$
$3.78 \times 10^{- 7} m, 5.60 eV$
$3.78 \times 10^{- 7} m, 1.20 eV$

1

2

3

4

(1)

By Einstein's photoelectric equation,

$\frac{h C}{λ} - ϕ = e V_{stopping}$

For first source, $\frac{h C}{λ} - ϕ = 6 e$ $\dots (i)$

For second source, $\frac{h C}{4 λ} - ϕ = 0.6 e$ $\dots (i i)$

Subtracting (ii) from (i), we get

$\frac{3 h C}{4 λ} = 5.4 e$

$λ = \frac{3}{4 \times 5.4} \times \frac{h C}{e} = \frac{3}{21.6} \times 1.24 \times 10^{- 6} = 1.72 \times 10^{- 7} m$

So, $ϕ = \frac{h C}{λ} - 6 e$ $[from (i)]$

$\Rightarrow ϕ (eV) = \frac{12400}{1720} - 6 = 1.2 eV$

Q 2 :

This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that describes the two statements.

Statement 1: Davisson–Germer experiment established the wave nature of electrons.

Statement 2: If electrons have wave nature, they can interfere and show diffraction. [2012]

Statement 1 is false, Statement 2 is true.
Statement 1 is true, Statement 2 is false.
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1.
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1.

1

2

3

4

(1)

Davisson-Germer experiment showed that electron beams can undergo diffraction when passed through an atomic crystal. This established the wave nature of electrons, as waves can exhibit interference and diffraction.

Q 3 :

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ( $λ$ ) of incident light and the corresponding stopping potential ( $V_{0}$ ) are given below:

$λ (μ m)$ $V_{0} (Volt)$

0.3 2.0

0.4 1.0

0.5 0.4

Given that $c = 3 \times 10^{8} {ms}^{- 1}$ and $e = 1.6 \times 10^{- 19} C,$ Planck's constant (in units of J s) found from such an experiment is [2016]

$6.0 \times 10^{- 34}$
$6.4 \times 10^{- 34}$
$6.6 \times 10^{- 34}$
$6.8 \times 10^{- 34}$

1

2

3

4

(2)

$\frac{h c}{e λ_{1}} - \frac{ϕ}{e} = V_{0_{1}}$ and $\frac{h c}{e λ_{2}} - \frac{ϕ}{e} = V_{0_{2}}$ $(∵ \frac{h c}{λ} - ϕ = e V_{0})$

$∴$ $\frac{h c}{e} [\frac{1}{λ_{1}} - \frac{1}{λ_{2}}] = V_{0_{1}} - V_{0_{2}}$

$∴$ $h = \frac{e (V_{0_{1}} - V_{0_{2}}) λ_{1} λ_{2}}{(λ_{2} - λ_{1}) c}$

From the first two values given in data,

$h = \frac{1.6 \times 10^{- 19} (2 - 1) \times 0.4 \times 0.3 \times 10^{- 6}}{0.1 \times 3 \times 10^{8}}$

$h = 0.64 \times 10^{- 33} = 6.4 \times 10^{- 34} J-s$

Similarly, if we calculate $h$ for the last two values of data, we get the same value of $h = 6.4 \times 10^{- 34} J-s$

Q 4 :

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_{1}$ and $u_{2}$ , respectively. If the ratio $u_{1} : u_{2} = 2 : 1,$ and $h c = 1240 eV nm,$ the work function of the metal is nearly [2014]

3.7 eV
3.2 eV
2.8 eV
2.5 eV

1

2

3

4

(1)

Here, $\frac{h C}{λ_{1}} - ϕ = \frac{1}{2} m u_{1}^{2}$ $\dots (i)$

and $\frac{h C}{λ_{2}} - ϕ = \frac{1}{2} m u_{2}^{2}$ $\dots (i i)$

Dividing equation (i) by (ii),

$\frac{\frac{h C}{λ_{1}} - ϕ}{\frac{h C}{λ_{2}} - ϕ} = \frac{u_{1}^{2}}{u_{2}^{2}}$

$∴$ $\frac{\frac{1240}{248} - ϕ}{\frac{1240}{310} - ϕ} = \frac{4}{1}$

$∴$ $\frac{1240}{248} - ϕ = 4 \times \frac{1240}{310} - 4 ϕ$

$∴$ $ϕ = 3.7 eV$

Hence, the work function of the metal is nearly, $ϕ = 3.7 eV$

Q 5 :

If $λ_{Cu}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{Mo}$ is the wavelength of the $K_{α}$ X-ray line of molybdenum (atomic number 42), then the ratio $\frac{λ_{Cu}}{λ_{Mo}}$ is close to [2014]

1.99
2.14
0.50
0.48

1

2

3

4

(2)

$∵$ $\frac{1}{λ} = R (z - b^{2}) (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

For K-series, $b = 1$ and for $K_{\propto}$ transition $n_{f} = 2$ to $n_{i} = 1$ ,

$∴$ $\frac{1}{λ} \propto {(z - 1)}^{2}$

$∵$ $\frac{λ_{Cu}}{λ_{Mo}} = \frac{{(Z_{Mo} - 1)}^{2}}{{(Z_{Cu} - 1)}^{2}} = {(\frac{42 - 1}{29 - 1})}^{2} = {(\frac{41}{28})}^{2} = 2.14$

Q 6 :

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is $3 \times 10^{8} {ms}^{- 1}$ . The final momentum of the object is [2013]

$0.3 \times 10^{- 17} kg {ms}^{- 1}$
$3.0 \times 10^{- 17} kg {ms}^{- 1}$
$1.0 \times 10^{- 17} kg {ms}^{- 1}$
$9.0 \times 10^{- 17} kg {ms}^{- 1}$

1

2

3

4

(3)

Final momentum,

$p = \frac{E}{c} = \frac{P \times t}{c} = \frac{30 \times 10^{- 3} \times 100 \times 10^{- 9}}{3 \times 10^{8}}$

$= 1.0 \times 10^{- 17} kg {ms}^{- 1}$

Q 7 :

Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? [2008]

Wavelength of characteristic X-rays decreases when the atomic number of the target increases
Cut-off wavelength of the continuous X-rays depends on the atomic number of the target
Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube

1

2

3

4

(2)

The wavelength of continuous X-rays is independent of the atomic number of the target material. The wavelength of characteristic X-rays depends on the atomic number of the target material.

Q 8 :

Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

$λ_{0} = \frac{2 m c λ^{2}}{h}$
$λ_{0} = \frac{2 h}{m c}$
$λ_{0} = \frac{2 m^{2} c^{2} λ^{3}}{h^{2}}$
$λ_{0} = λ$

1

2

3

4

(1)

The cut-off wavelength of the emitted X-ray,

$λ_{0} = \frac{h c}{e V}$ $\dots (i)$

According to de Broglie equation,

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}$

or $λ^{2} = \frac{h^{2}}{2 m e V}$ $\Rightarrow V = \frac{h^{2}}{2 m e λ^{2}}$ $\dots (i i)$

From eq. (i) and (ii),

$λ_{0} = \frac{h c \times 2 m e λ^{2}}{e h^{2}} = \frac{2 m c λ^{2}}{h}$

Q 9 :

$K_{α}$ wavelength emitted by an atom of atomic number Z = 11 is $λ$ . Find the atomic number for an atom that emits $K_{α}$ radiation with wavelength $4 λ$ . [2005]

Z = 6
Z = 4
Z = 11
Z = 44

1

2

3

4

(1)

$For K_{α}, \frac{1}{λ} \propto {(Z - 1)}^{2}$

$∴$ $\frac{λ_{2}}{λ_{1}} = \frac{{(Z_{1} - 1)}^{2}}{{(Z_{2} - 1)}^{2}}$ $\Rightarrow \frac{4 λ}{λ} = \frac{{(11 - 1)}^{2}}{{(Z_{2} - 1)}^{2}}$

$\Rightarrow Z_{2} - 1 = \frac{10}{2}$

$∴$ $Z_{2} = 6$

Q 10 :

In a photoelectric experiment anode potential is plotted against plate current. [2004]

[IMAGE 1177]

A and B will have different intensities while B and C will have different frequencies.
B and C will have different intensities while A and C will have different frequencies.
A and B will have different intensities while A and C will have equal frequencies.
B and C will have equal intensities while A and B will have same frequencies.

1

2

3

4

(4)

Saturation current $\propto$ (intensity) and stopping potential increase with increase in frequency. From the graph it is clear that A and B have the same stopping potential and therefore, the same frequency. Also, B and C have the same intensity.

Q 11 :

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 $m A$ . Then the number of electrons striking the target per second is [2002]

$2 \times 10^{16}$
$5 \times 10^{6}$
$1 \times 10^{17}$
$4 \times 10^{15}$

1

2

3

4

(1)

As we know, $I = \frac{q}{t} = \frac{n e}{t}$

$∴$ No. of electrons striking the target per second

$\frac{n}{t} = \frac{I}{e} = 2 \times 10^{16}$

Q 12 :

The intensity of X-rays from a Coolidge tube is plotted against wavelength $λ$ as shown in the figure. The minimum wavelength found is $λ_{C}$ and the wavelength of the $K_{α}$ line is $λ_{K}$ . As the accelerating voltage is increased [2001]

[IMAGE 1178]

$λ_{K} - λ_{C}$ increases
$λ_{K} - λ_{C}$ decreases
$λ_{K}$ increases
$λ_{K}$ decreases

1

2

3

4

(1)

In case of Coolidge tube, $λ_{\min} = \frac{h c}{e V} = λ_{c}$

Thus the minimum wavelength is inversely proportional to accelerating voltage. As $V$ increases, $λ_{c}$ decreases. $λ_{k}$ is the wavelength of $K_{α}$ line which is independent of accelerating voltage of bombarding electron. Since $λ_{k}$ always refers to a photon wavelength of transition of $e^{-}$ from the target element from $2 \to 1$ . Therefore $λ_{k} - λ_{c}$ increases as accelerating voltage is increased.

Q 13 :

Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only [2000]

a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 $Å$
a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
the characteristic X-ray spectrum of tungsten
a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 $Å$ and the characteristic X-ray spectrum of tungsten.

1

2

3

4

(4)

Minimum wavelength of continuous X-ray spectrum

$λ_{\min} = \frac{h c}{E}$

$∴$ $λ_{\min} = \frac{12375}{80 \times 10^{3}} Å = 0.155 Å$ $[∵ E = 80 keV given]$

Energy of incident electrons is greater than the ionization energy of electrons in the K-shell, the K-shell electrons will be knocked off. Hence, characteristic X-ray spectrum will be obtained.

Q 14 :

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_{P}$ , $E_{Q}$ and $E_{R}$ , respectively, and they are related by $E_{P} = 2 E_{Q} = 2 E_{R} .$ In this experiment, the same source of monochromatic light is used for metals $P$ and $Q$ while a different source of monochromatic light is used for the metal $R$ . The work functions for metals $P, Q$ and $R$ are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal $R$ , in eV, is ________. [2021]

(6)

From photoelectric equation, $E = ϕ_{0} + {K . E .}_{\max}$

For metals $P$ and $Q$ ,

$E_{1} = ϕ_{P} + {KE}_{P} \Rightarrow E_{1} - 4 = E_{P}$ and $E_{1} - 4.5 = E_{Q}$

$E_{P} = 2 E_{Q}$ $(given)$

$∴$ $E_{1} - 4 = 2 (E_{1} - 4.5) \Rightarrow E_{1} = 5 eV$

$E_{1} - 4 = E_{P} \Rightarrow 5 - 4 = E_{P}$

$∴$ $E_{P} = 1 eV$ and $E_{Q} = E_{R} = 0.5 eV$

For metal $R$ , $E_{2} - ϕ_{R} = {KE}_{R}$

$∴$ $E_{2} - 5.5 = 0.5$

$∴$ $E_{2} = 0.5 + 5.5 = 6 eV$

Q 15 :

A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency such that $= 10^{24} m^{- 2},$ with $h$ as Planck's constant. $N$ photons of wavelength $λ = 8 π \times 10^{- 6} m$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by $1 μ m$ . If the value of $N$ is $x \times 10^{12}$ , then the value of $x$ is ________. [Consider the spring as massless] [2019]

[IMAGE 1179]

(1)

From conservation of momentum principle, change in momentum of photon = change in momentum of mirror

$2 (N P) = M V_{\max}$

$\Rightarrow 2 [N (\frac{h}{λ})] = M V_{m a x}$

$∴$ $2 \frac{N h}{λ} = M (A Ω)$ $[∵ V_{\max} = A Ω]$

$N = (\frac{M Ω}{h}) \frac{A λ}{2} = \frac{10^{24}}{4 π} \times \frac{10^{- 6} \times 8 π \times 10^{- 6}}{2}$ $[∵ \frac{M Ω}{h} = \frac{10^{24}}{4 π}, A = 1 hm, λ = 8 π \times 10^{- 6}]$

$∴$ $N = 1 \times 10^{12} = x \times 10^{12}$

$∴$ $x = 1$

Q 16 :

The work functions of Silver and Sodium are 4.6 eV and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is _____. [2013]

(1)

[IMAGE 1180]

For photoelectric effect

$h ν - ϕ = e V_{0}$

$\Rightarrow \frac{h ν}{e} - \frac{ϕ_{0}}{e} = V_{0}$

The slope of $V_{0}$ versus $ν$ graph is a straight line with slope

$\tan θ = \frac{h}{e} = constant$

$∴$ The ratio of two slopes will be 1.

Q 17 :

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^{z}$ (where $1 < A < 10$ ). The value of $' z'$ is _____. [2011]

(7)

From $\frac{h c}{λ} - ϕ = - e V_{0}$

Stopping potential $V_{0} = \frac{1}{e} [\frac{h c}{λ} - ϕ]$

where $h c = 1240 eV - nm$

$= \frac{1}{e} [\frac{1240}{200} - 4.7] = \frac{1}{e} [6.2 - 4.7] = \frac{1}{e} \times 1.5 eV = 1.5 V$

But $V = \frac{1}{4 π ε_{0}} \frac{q}{r} = \frac{1}{4 π ε_{0}} \frac{n e}{r}$ $(n = no. of photo-electrons emitted)$

$\Rightarrow n = \frac{V r (4 π ε_{0})}{e} = \frac{1.5 \times 10^{- 2}}{9 \times 10^{9} \times 1.6 \times 10^{- 19}}$

$∴$ $n = 1.04 \times 10^{7}$

Comparing it with $A \times 10^{z}$ , we get, $z = 7$

Q 18 :

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $ν_{1}$ and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $ν_{2}$ . The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that the positron has the same mass as that of the electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is _____. [2025]

(11.80)

$E_{1} = h ν_{1} = 13.6 + 10 = 23.6 eV$

Energy of positronium in ground state

$= - 13.6 \frac{μ}{m} {(\frac{Z}{n})}^{2} eV = - 13.6 \times \frac{1}{2} eV = - 6.8 eV$

$E_{2} = h ν_{2} = (10 - 5) + 6.8 = 11.8 eV$

$∴$ $Difference in energy = 23.6 - 11.8 = 11.8 eV$

Q 19 :

A cube of unit volume contains $35 \times 10^{7}$ photons of frequency $10^{15} Hz$ . If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $α \times 10^{- 9} T$ . Taking permeability of free space $μ_{0} = 4 π \times 10^{- 7} Tm / A,$ Planck's constant $h = 6 \times 10^{- 34} Js$ and $π = \frac{22}{7},$ the value of $α$ is ______. [2025]

(22.98)

$Total energy in cube$ $E = n h f = 35 \times 10^{7} \times h f$

$= 35 \times 10^{7} \times 6 \times 10^{- 34} \times 10^{15} = 2.1 \times 10^{- 10} J$

$Total energy of EM waves = \frac{B_{0}^{2}}{2 μ_{0}} \times volume$

$B_{0}^{2} = \frac{2.1 \times 10^{- 10} \times 8 π \times 10^{- 7}}{1^{3}} = 22.98 \times 10^{- 9} T$

$∴$ $α = 22.98$

Q 20 :

In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n \times 10^{- 4} N$ due to the impact of the electrons. The value of $n$ is ______. Mass of the electron $m_{e} = 9 \times 10^{- 31} kg$ and $1.0 eV = 1.6 \times 10^{- 19} J .$ [2018]

(24)

$Number of electrons emitted/s$

$= \frac{200 W}{6.25 \times 1.6 \times 10^{- 19} J}$

$Force,$ $F = Rate of change of linear momentum = N \sqrt{2 m V}$

$= \frac{200}{6.25 \times 1.6 \times 10^{- 19}} \times \sqrt{2 \times 9 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times 500}$

$\Rightarrow F = 24 \times 10^{- 4} N$ $[∵ K = e V, e = 1.6 \times 10^{- 19}, V = 500]$

$∴$ $n = 24.00$

Q 21 :

Light of wavelength $λ_{ph}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $ϕ$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. A potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $λ_{e}$ , which of the following statement(s) is (are) true? [2016]

[IMAGE 1181]

$λ_{e}$ decreases with increase in $ϕ$ and $λ_{ph}$
$λ_{e}$ is approximately halved, if $d$ is doubled
For large potential difference $(V ≫ ϕ / e)$ , $λ_{e}$ is approximately halved if $V$ is made four times
$λ_{e}$ increases at the same rate as $λ_{ph}$ for $λ_{ph} < \frac{h c}{ϕ}$

1

2

3

4

(3)

de-Broglie wavelength passing through the anode,

$λ_{e} = \frac{h}{p} = \frac{h}{\sqrt{2 m (K . E .)}}$

When $ϕ$ increases, K.E. decreases and therefore $λ_{e}$ increases.

When $λ_{p h}$ increases, $N_{p h}$ decreases, K.E. decreases and therefore $λ_{e}$ increases.

$λ_{e}$ is independent of the distance $d$ .

Also after reaching anode, $\frac{h c}{λ_{p h}} + e V - ϕ = \frac{h^{2}}{2 m λ_{e}^{2}}$

$[λ_{e} = \frac{h}{\sqrt{2 m (K . E .)}}]$ $∴$ $\frac{h c}{e λ_{p h}} + V - \frac{ϕ}{e} = \frac{h^{2}}{2 m e λ_{e}^{2}}$ $\dots (i)$

For $V ≫ \frac{ϕ}{e}, ϕ ≪ e V$

Also $\frac{h c}{e λ_{p h}} ≪ V$ $∴$ $from eq. (i), λ_{e} \propto \frac{1}{\sqrt{V}}$

Hence, if $V$ is made four times, $λ_{e}$ is approximately half.

Q 22 :

For photo-electric effect with incident photon wavelength $λ$ , the stopping potential is $V_{0}$ . Identify the correct variation(s) of $V_{0}$ with $λ$ and $\frac{1}{λ}$ . [2015]

[IMAGE 1182]
[IMAGE 1183]
[IMAGE 1184]
[IMAGE 1185]

1

2

3

4

Select one or more options

(1, 3)

We know that

$\frac{h c}{λ} - ϕ = e V_{0} \Rightarrow \frac{h c}{e λ} - \frac{ϕ}{e} = V_{0} \Rightarrow \frac{h c}{e} (\frac{1}{λ}) - \frac{ϕ}{e} = V_{0}$

Therefore, $V_{0}$ versus $\frac{1}{λ}$ graph is a straight line with negative slope $(\frac{h c}{e})$ and positive intercept $(\frac{ϕ}{e}) .$

For $V_{0}$ versus $λ$ , we will get a hyperbola. As $λ$ decreases, $V_{0}$ increases. And $V_{0}$ becomes zero when $(\frac{h c}{λ}) = ϕ$ i.e., when $λ = λ_{0}$

Q 23 :

The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct? [2006]

[IMAGE 1186]

$ϕ_{1} : ϕ_{2} : ϕ_{3} = 1 : 2 : 4$
$ϕ_{1} : ϕ_{2} : ϕ_{3} = 4 : 2 : 1$
$\tan θ \propto \frac{h c}{e}$
Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

1

2

3

4

Select one or more options

(1, 3)

From $e V = \frac{h c}{λ} - ϕ$

$V = \frac{h c}{e λ} - \frac{ϕ}{e}$

At $V = 0$ , $ϕ_{1} : ϕ_{2} : ϕ_{3} = \frac{h c}{λ_{1}} : \frac{h c}{λ_{2}} : \frac{h c}{λ_{3}}$

$= 0.001 : 0.002 : 0.004 = 1 : 2 : 4$

[IMAGE 1187]

By Einstein's photoelectric equation, $\frac{h c}{λ} - ϕ = e V$

$\Rightarrow V = \frac{h c}{e λ} - \frac{ϕ}{e}$ $\dots (i)$

Comparing equation (i) by $y = m x + c,$ we get the slope of the line, $m = \frac{h c}{e} = \tan θ$

From the graph of $V_{0}$ versus $\frac{1}{λ}$ , it is clear that,

$\frac{1}{λ_{0_{1}}} = 0.001 {(nm)}^{- 1}$ $∴$ $λ_{0_{1}} = \frac{1}{0.001} = 1000 nm$

$\frac{1}{λ_{0_{2}}} = 0.002 {(nm)}^{- 1}$ $∴$ $λ_{0_{2}} = 500 nm$ and $λ_{0_{3}} = 250 nm$

Violet colour light has wavelength 400 nm.

Therefore, this light will be unable to show photoelectric effect on plate 3 and can eject photoelectrons from metal-1 and metal-2.

Q 24 :

Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option. [2023]

[Given: Wien's constant as $2.9 \times 10^{- 3} m K$ and $\frac{h c}{e} = 1.24 \times 10^{- 6} V m$ ]

List-I List-II

(P) 2000 K (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV.

(Q) 3000 K (2) The radiation at peak wavelength is visible to human eye.

(R) 5000 K (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.

(S) 10000 K (4) The power emitted per unit area is 1/16 of that emitted by a blackbody at temperature 6000 K.

(5) The radiation at peak emission wavelength can be used to image human bones.

P → 3, Q → 5, R → 2, S → 3
P → 3, Q → 2, R → 4, S → 1
P → 3, Q → 4, R → 2, S → 1
P → 1, Q → 2, R → 5, S → 3

1

2

3

4

(3)

From Wien's displacement law, $λ_{m} T = b (constant)$

and $b = 2.9 \times 10^{- 3} mK$

When $T = 2000 K$ (P), $λ_{m} = \frac{b}{T} = \frac{2.9 \times 10^{- 3}}{2000} = 1450 nm (\max)$

When $T = 3000 K$ (Q), $λ_{m} = \frac{2.9 \times 10^{- 3}}{3000} = 966.66 nm$

When $T = 5000 K$ (R), $λ_{m} = \frac{2.9 \times 10^{- 3}}{5000} = 580 nm$

When $T = 10000 K$ (S), $λ_{m} = \frac{2.9 \times 10^{- 3}}{10000} = 290 nm (\min)$

For option (P), $λ$ maximum and $β = \frac{λ D}{d}$

$∴$ widest central maximum

For option (Q), Power

$P_{3000} = σ A {(3000)}^{4}$

$P_{6000} = σ A {(6000)}^{4}$

$\frac{P_{3000}}{P_{6000}} = {(\frac{1}{2})}^{4} = \frac{1}{16}$

$∴$ $P_{3000} = \frac{1}{16} P_{6000}$

For option (R), Wavelength $λ = 580 nm$

Visible to human eyes.

For (S), $λ = \frac{h c}{ϕ} = \frac{1.24 \times 10^{- 6}}{4} = 310 nm$

$290 nm < 310 nm$ so this radiation leads to emission of photoelectrons from a metal of work function $ϕ = 4 eV .$

For imaging human bones, X-rays of wavelength range 1 - 10 nm are used.

Q 25 :

STATEMENT-1: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.

STATEMENT-2: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. [2007]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is False.
Statement-1 is False, Statement-2 is True.

1

2

3

4

(2)

Note: Shortest wavelength means highest frequency, so highest energy.

Shortest or cut-off wavelength of X-rays emitted from an X-ray tube depends on the voltage applied to the tube.

Also, according to Moseley's law, $\sqrt{ν} = a (Z - b) .$ Thus, the frequency also depends on the atomic number.

Topic Question Set

Q 5 :

If $λ_{Cu}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{Mo}$ is the wavelength of the $K_{α}$ X-ray line of molybdenum (atomic number 42), then the ratio $\frac{λ_{Cu}}{λ_{Mo}}$ is close to [2014]

Q 6 :

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is $3 \times 10^{8} {ms}^{- 1}$ . The final momentum of the object is [2013]

Q 7 :

Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? [2008]

Q 8 :

Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

Q 9 :

$K_{α}$ wavelength emitted by an atom of atomic number Z = 11 is $λ$ . Find the atomic number for an atom that emits $K_{α}$ radiation with wavelength $4 λ$ . [2005]

Q 10 :

In a photoelectric experiment anode potential is plotted against plate current. [2004]

[IMAGE 1177]

Q 11 :

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 $m A$ . Then the number of electrons striking the target per second is [2002]

Q 12 :

The intensity of X-rays from a Coolidge tube is plotted against wavelength $λ$ as shown in the figure. The minimum wavelength found is $λ_{C}$ and the wavelength of the $K_{α}$ line is $λ_{K}$ . As the accelerating voltage is increased [2001]

[IMAGE 1178]

Q 13 :

Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only [2000]

Q 16 :

The work functions of Silver and Sodium are 4.6 eV and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is _____. [2013]

Q 22 :

For photo-electric effect with incident photon wavelength $λ$ , the stopping potential is $V_{0}$ . Identify the correct variation(s) of $V_{0}$ with $λ$ and $\frac{1}{λ}$ . [2015]

Q 23 :

The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct? [2006]

[IMAGE 1186]

Q 25 :

STATEMENT-1: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.

STATEMENT-2: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. [2007]

Want Us to Email you About Special Offers & Updates?

	List-I		List-II
(P)	2000 K	(1)	The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV.
(Q)	3000 K	(2)	The radiation at peak wavelength is visible to human eye.
(R)	5000 K	(3)	The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
(S)	10000 K	(4)	The power emitted per unit area is 1/16 of that emitted by a blackbody at temperature 6000 K.
		(5)	The radiation at peak emission wavelength can be used to image human bones.

Topic Question Set

Q 4 : A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1:u2=2:1, and hc=1240 eV nm, the work function of the metal is nearly [2014]

Q 5 : If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCuλMo is close to [2014]

Q 6 : A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×108 ms-1. The final momentum of the object is [2013]

Q 7 : Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? [2008]

Q 8 : Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

Q 9 : Kα wavelength emitted by an atom of atomic number Z = 11 is λ. Find the atomic number for an atom that emits Kα radiation with wavelength 4λ. [2005]

Q 10 : In a photoelectric experiment anode potential is plotted against plate current. [2004] [IMAGE 1177]

Q 11 : The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electrons striking the target per second is [2002]

Q 12 : The intensity of X-rays from a Coolidge tube is plotted against wavelength λ as shown in the figure. The minimum wavelength found is λC and the wavelength of the Kα line is λK. As the accelerating voltage is increased [2001] [IMAGE 1178]

Q 13 : Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only [2000]

Q 16 : The work functions of Silver and Sodium are 4.6 eV and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is _____. [2013]

Q 22 : For photo-electric effect with incident photon wavelength λ, the stopping potential is V0. Identify the correct variation(s) of V0 with λ and 1λ. [2015]

Q 23 : The graph between the stopping potential (V0) and (1λ) is shown in the figure. ϕ1, ϕ2 and ϕ3 are work functions, which of the following is/are correct? [2006] [IMAGE 1186]

Q 25 : STATEMENT-1: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. STATEMENT-2: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. [2007]

Want Us to Email you About Special Offers & Updates?

Q 5 :

If $λ_{Cu}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{Mo}$ is the wavelength of the $K_{α}$ X-ray line of molybdenum (atomic number 42), then the ratio $\frac{λ_{Cu}}{λ_{Mo}}$ is close to [2014]

Q 6 :

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is $3 \times 10^{8} {ms}^{- 1}$ . The final momentum of the object is [2013]

Q 7 :

Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube? [2008]

Q 8 :

Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

Q 9 :

$K_{α}$ wavelength emitted by an atom of atomic number Z = 11 is $λ$ . Find the atomic number for an atom that emits $K_{α}$ radiation with wavelength $4 λ$ . [2005]

Q 10 :

In a photoelectric experiment anode potential is plotted against plate current. [2004]

[IMAGE 1177]

Q 11 :

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 $m A$ . Then the number of electrons striking the target per second is [2002]

Q 12 :

The intensity of X-rays from a Coolidge tube is plotted against wavelength $λ$ as shown in the figure. The minimum wavelength found is $λ_{C}$ and the wavelength of the $K_{α}$ line is $λ_{K}$ . As the accelerating voltage is increased [2001]

[IMAGE 1178]

Q 13 :

Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only [2000]

Q 16 :

The work functions of Silver and Sodium are 4.6 eV and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is _____. [2013]

Q 22 :

For photo-electric effect with incident photon wavelength $λ$ , the stopping potential is $V_{0}$ . Identify the correct variation(s) of $V_{0}$ with $λ$ and $\frac{1}{λ}$ . [2015]

Q 23 :

The graph between the stopping potential $(V_{0})$ and $(\frac{1}{λ})$ is shown in the figure. $ϕ_{1}$ , $ϕ_{2}$ and $ϕ_{3}$ are work functions, which of the following is/are correct? [2006]

[IMAGE 1186]

Q 25 :

STATEMENT-1: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change.

STATEMENT-2: When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. [2007]