# A monochromatic neon

Question:

A monochromatic neon lamp with wavelength of $670.5 \mathrm{~nm}$ illuminates a photo-sensitive material which has a stopping voltage of $0.48 \mathrm{~V}$. What will be the stopping voltage if the source light is changed with another source of wavelength of $474.6 \mathrm{~nm}$ ?

1. $0.96 \mathrm{~V}$

2. $1.25 \mathrm{~V}$

3. $0.24 \mathrm{~V}$

4. $1.5 \mathrm{~V}$

Correct Option: , 2

Solution:

$\mathrm{kE}_{\max }=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}+\phi$

or $\quad \mathrm{eV}_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}+\phi$

when $\lambda_{\mathrm{i}}=670.5 \mathrm{~nm} ; \quad \mathrm{V}_{0}=0.48$

when $\lambda_{\mathrm{i}}=474.6 \mathrm{~nm} ; \quad \mathrm{V}_{\mathrm{o}}=$ ?

So, $\quad \mathrm{e}(0.48)=\frac{1240}{670.5}+\phi$...(1)

$e\left(V_{0}\right)=\frac{1240}{474.6}+\phi$...(2)

.$(2)-(1)$

$e\left(V_{0}-0.48\right)=1240\left(\frac{1}{474.6}-\frac{1}{670.5}\right) e V$

$V_{o}=0.48+1240\left(\frac{670.5-474.6}{474.6 \times 670.5}\right)$ Volts

$\mathrm{V}_{\mathrm{o}}=0.48+0.76$

$\mathrm{~V}_{\mathrm{o}}=1.24 \mathrm{~V} \simeq 1.25 \mathrm{~V}$