Question

In an experiment of photoelectric effect, the stopping potential was measured to be $V_{1}$ and $V_{2}$ volts with incident light of wavelength $\lambda$ and $\lambda / 2$ respectively. The relation between $V_{1}$ and $V_{2}$ may be

• A. $V_{2} < V_{1}$
• B. $V_{1} < V_{2} < 2 V_{1}$
• C. $V_{2} = 2 V_{1}$
• D. $V_{2} > 2 V_{1}$

Answer 1

Answer: (D)

According to Einstein's photoelectric equation
$K_{\max }=h v-\phi_{0} $
$e V_{s}=\frac{h c}{\lambda}-\phi_{0} $
$\Rightarrow V_{s}=\frac{h c}{\lambda e}-\frac{\phi_{0}}{e}$
where, $\lambda=$ Wavelength of incident light
$\phi_{0}=$ Work function
$V_{s}=$ Stopping potential
According to given problem
$V_{1}=\frac{h c}{\lambda e}-\frac{\phi_{0}}{e} \ldots( i )$
and $ V_{2}=\frac{h c}{\left(\frac{\lambda}{2}\right) e}-\frac{\phi_{0}}{e} \dots$(ii)
$V_{2}=\frac{2 h c}{\lambda e}-\frac{\phi_{0}}{e}$
$=\frac{2 h c}{\lambda e}-\frac{2 \phi_{0}}{e}+\frac{2 \phi_{0}}{e}-\frac{\phi_{0}}{e}$
$=2\left(\frac{h c}{\lambda e}-\frac{\phi_{0}}{e}\right)+\frac{\phi_{0}}{e}$
$V_{2}=2 V_{1}+\phi_{0} $
$\therefore V_{2} > 2 V_{1} $ (Using (i))