SATHEE: Modern Physics 4 Question 8

Modern Physics 4 Question 8

8. A photoelectric material having work-function $φ_{O}$ is illuminated with light of wavelength $λ λ < \frac{h c}{φ_{0}}$ . The fastest photoelectron has a de-Broglie wavelength $λ_{d}$ . A change in wavelength of the incident light by $Δ λ$ results in a change $Δ λ_{d}$ in $λ_{d}$ . Then, the ratio $\frac{Δ λ_{d}}{Δ λ}$ is proportional to

(a) $\frac{λ_{d}^{2}}{λ^{2}}$

(b) $\frac{λ_{d}}{λ}$

(d) $\frac{λ_{d}^{3}}{λ^{2}}$

(2017 Adv.)

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Solution:

According to photoelectric effect equation

$\begin{aligned} K E_{max} & = \frac{h c}{λ} - φ_{0} \\ \frac{p^{2}}{2 m} & = \frac{h c}{λ} - φ_{0} & [K E = p^{2} / 2 m] \\ \frac{{(h / λ_{d})}^{2}}{2 m} & = \frac{h c}{λ} - φ_{0} & [p = h / λ] \end{aligned}$

Assuming small changes, differentiating both sides,

$\frac{h^{2}}{2 m} - \frac{2 d λ_{d}}{λ_{d}^{3}} = - \frac{h c}{λ^{2}} d λ, \frac{d λ_{d}}{d λ} \propto \frac{λ_{d}^{3}}{λ^{2}}$

Modern Physics 4 Question 8

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Modern Physics 4 Question 8

8. A photoelectric material having work-function φO is illuminated with light of wavelength λλ<hcφ0. The fastest photoelectron has a de-Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then, the ratio ΔλdΔλ is proportional to

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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