Dual Nature Of Radiation And Matter | Physics JEE previous year questions with Solutions

Q 31 :

A metallic surface is illuminated with radiation of wavelength $λ$ , the stopping potential is $V_{0}$ . If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential becomes $\frac{V_{0}}{4}$ . The threshold wavelength for this metallic surface will be [2023]

$\frac{λ}{4}$
$4 λ$
$\frac{3}{2} λ$
$3 λ$

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(4)

$From the equation of photoelectric effect$

$e V_{0} = \frac{h c}{λ} - ϕ_{0} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$\frac{e V_{0}}{4} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\Rightarrow \frac{1}{4} (\frac{h c}{λ} - \frac{h c}{λ_{0}}) = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\frac{1}{λ_{0}} - \frac{1}{4 λ_{0}} = \frac{1}{2 λ} - \frac{1}{4 λ}$

$\frac{3}{4 λ_{0}} = \frac{1}{4 λ} \Rightarrow λ_{0} = 3 λ$

Q 32 :

The difference between threshold wavelengths for two metal surfaces A and B having work function $ϕ_{A} = 9 eV$ and $ϕ_{B} = 4.5 eV$ in nm is: (Given, $h c = 1242 eV nm$ ) [2023]

264
138
276
540

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(2)

$λ_{A} = (\frac{1242}{9}) = 138 nm$

$λ_{B} = (\frac{1242}{4.5}) = 276 nm$

$λ_{B} - λ_{A} = 138 nm$

Q 33 :

Given below are two statements: [2023]

Statement I: Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II: Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer from the options given below:

Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Both Statement I and Statement II are false

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(1)

UV rays have maximum frequency hence are most effective for emission of electrons from a metallic surface.

$K E_{\max} = h f - h f_{0}$

Q 34 :

A light wave described by $E = 60 [\sin (3 \times 10^{15}) t + \sin (12 \times 10^{15}) t]$ (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ______ eV. $(h = 6.6 \times 10^{- 34} J·s and e = 1.6 \times 10^{- 19} C)$ [2026]

7.8
3.8
5.1
6.0

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(3)

$ω_{1} = 3 \times 10^{15} rad/sec$

$ω_{2} = 12 \times 10^{15} rad/sec$

$∵ ν = \frac{ω}{2 π}$

$E_{photon} = h ν = 6.6 \times 10^{- 34} \times 1.91 \times 10^{15}$

$= 1.26 \times 10^{- 18} J$

$E_{\max} = \frac{1.26 \times 10^{- 18}}{1.6 \times 10^{- 19}} \approx 7.9 eV$

$K_{\max} = E_{\max} - ϕ_{0}$

$= 7.9 - 2.8$

$K_{\max} = 5.1 eV$

Q 35 :

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is _______ m.

$(h = 6.63 \times 10^{- 34} J s, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} m/s)$ [2026]

$2.2 \times 10^{- 8}$
$3.1 \times 10^{- 7}$
$2.9 \times 10^{- 8}$
$2.5 \times 10^{- 7}$

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Q 36 :

Light is incident on a metallic plate having work function $110 \times 10^{- 20} J .$ If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s.
$(h = 6.63 \times 10^{- 34} J·s)$ [2026]

$1.66 \times 10^{15}$
$1.04 \times 10^{16}$
$1.04 \times 10^{13}$
$1.66 \times 10^{16}$

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(2)

$ϕ = h ν$

$ν = \frac{ϕ}{h}$

$ω = 2 π ν = \frac{2 π ϕ}{h} = \frac{2 \times 3.14 \times 110 \times 10^{- 2}}{6.63 \times 10^{- 34}}$

$ω = 1.04 \times 10^{16} rad/sec$

Topic Question Set

Q 32 :

The difference between threshold wavelengths for two metal surfaces A and B having work function $ϕ_{A} = 9 eV$ and $ϕ_{B} = 4.5 eV$ in nm is: (Given, $h c = 1242 eV nm$ ) [2023]

Q 36 :

Light is incident on a metallic plate having work function $110 \times 10^{- 20} J .$ If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s.
$(h = 6.63 \times 10^{- 34} J·s)$ [2026]

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