Question

A photoelectric material having work-function ${\varphi }_0$ is illuminated with light of wavelength $\lambda (\lambda <\frac{hc}{{\varphi }_0})$. The fastest photoelectron has a de-Broglie wavelength ${\lambda }_d$. A change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta {\lambda }_d$ in ${\lambda }_d$. Then the ratio $\frac{\Delta {\lambda }_d}{\Delta \lambda }$ is proportional to

• A. $\frac{{\lambda }_d^3}{{\lambda }^2}$
• B. $\frac{{\lambda }_d^3}{\lambda }$
• C. $\frac{{\lambda }_d^2}{{\lambda }^2}$
• D. $\frac{{\lambda }_d}{\lambda }$

Answer 1

The correct option is A $\frac{{\lambda }_d^3}{{\lambda }^2}$

From, Einstein's photo electric equation,

$E={\varphi }_0+K{E}_{max}$

$\frac{hc}{\lambda }={\varphi }_0+K{E}_{max}$

$\because K{E}_e=\frac{p^2}{2m_e}=\frac{h^2}{2m_e{\lambda }_d^2}[\because p=\frac{h}{\lambda }]$

$\Rightarrow \frac{hc}{\lambda }={\varphi }_0+\frac{h^2}{2m_e{\lambda }_d^2}$

Differentiating on both sides,

$-\frac{hc}{{\lambda }^2}d\lambda =0+\frac{h^2}{2m_e}\frac{(-2)}{{\lambda }_d^3}d{\lambda }_d$

$\Rightarrow \frac{d{\lambda }_d}{d\lambda }=\frac{m_e{\lambda }_d^3\times hc}{h^2\times {\lambda }^2}$

$\therefore \frac{d{\lambda }_d}{d\lambda }\propto \frac{{\lambda }_d^3}{{\lambda }^2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question? Key Note: In 1924 Louis de-Broglie (a French Scientist) presented the logic that, “A moving particle is always associated with a wave. Hence Photon also consists of the properties of waves. This concept of de-Broglie is also known as the dual nature of particles. Any moving particle i.e. electrons, photons are always associated with a wave. And these waves are called matter waves or de-Broglie waves. And the wavelengths of these waves are known as de-Broglie wavelengths.