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

Question

Dr. Abubacker · Accepted Answer

(d) According to Einstein’s photoelectric equation
$K_{\max} = hv - \phi_{0}$
$eV_{s} = \frac{hc}{\lambda} - \phi_{0} \Rightarrow V_{s} = \frac{hc}{\lambda e} - \frac{\phi_{0}}{e}$
Where, $\lambda =$ Wavelength of incident light
$\phi_{0} =$ Work function
$V_{s} =$ Stopping potential
According to given problem
$V_{1} = \frac{hc}{\lambda e} - \frac{\phi_{0}}{e}$
$V_{2} = \frac{hc}{\left( \frac{\lambda}{2} ight)e} - \frac{\phi_{0}}{e}$
$V_{2} = \frac{2hc}{\lambda e} - \frac{\phi_{0}}{e} = \frac{2hc}{\lambda e} - \frac{2\phi_{0}}{e} + \frac{2\phi_{0}}{e} - \frac{\phi_{0}}{e}$
$= 2\left( \frac{hc}{\lambda e} - \frac{\phi_{0}}{e} ight) + \frac{\phi_{0}}{e}$
$V_{2} = 2V_{1} + \frac{\phi_{0}}{e}$ [Using (i)]
$	herefore V_{2} > 2V_{1}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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