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 $d\lambda$, results in a change $d{\lambda }_d$ in ${\lambda }_d$. Then, the ratio $\frac{d{\lambda }_d}{d\lambda }$ is proportional to,

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

Answer 1

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

From photoelectric effect
we know $\frac{hc}{\lambda }=K.E_{\text{max}}+\varphi -\left(i\right)$

For an electron, $K.E=\frac{p^2}{2m_e}=\frac{{\left(\frac{h}{{\lambda }_d}\right)}^2}{2m_e}-\left(ii\right)$

Where ${\lambda }_d=\frac{h}{p},$ is the de-Broglie wavelength.

From $\left(i\right)$ & $\left(ii\right)$, we can write,

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

Differentiating with respect to $\lambda$,

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

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

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

Hence, $\left(A\right)$ is the correct answer.