Question

The wavelength of incident light falling on a photosensitive surface is changed from 2000 to 2050$A^0$. The corresponding change in the stopping potential is nearly

• A. $0.012V$
• B. $2.1V$
• C. $1.2V$
• D. $0.15V$

Answer 1

The correct option is D $0.15V$

According to given condition,

$\frac{hc}{2000}-W=E----\left(1\right)$

Where E $=eV$[ where V is stopping potential]

In the 2 nd case,

Change in energy of incident light,

$\frac{hc}{2000}-\frac{hc}{2050}=\left(\frac{hc}{{10}^4}\right)\times {10}^{10}\times 1.213512\times {10}^{-5}$

This change will be equal to change in stopping potential as

work function or W of the metal is fixed.

$\Rightarrow 10\times 1.219512\times {10}^{10}\times \frac{hc}{{10}^4}=eV\Rightarrow \frac{1.219512\times 10\times 6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}=V$

$\Rightarrow V=0.1516$ volts

$\Rightarrow V=0.15$ volts.