In the experiment of the photoelectric effect | vedclass

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Question

(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Since $K_{max} = eV_{s}$,w

Vedclass · Accepted Answer

$V_{0} - \frac{hc}{2e\lambda_{0}}$

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(B) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.Since $K_{max} = eV_{s}$,where $V_{s}$ is the stopping potential,we have $eV_{s} = \frac{hc}{\lambda} - \phi$.For the first case: $eV_{0} = \frac{hc}{\lambda_{0}} - \phi$  --- $(1)$For the second case with wavelength $2\lambda_{0}$: $eV_{s}' = \frac{hc}{2\lambda_{0}} - \phi$ --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$eV_{0} - eV_{s}' = \left(\frac{hc}{\lambda_{0}} - \phi\right) - \left(\frac{hc}{2\lambda_{0}} - \phi\right)$$e(V_{0} - V_{s}') = \frac{hc}{\lambda_{0}} - \frac{hc}{2\lambda_{0}} = \frac{hc}{2\lambda_{0}}$$V_{0} - V_{s}' = \frac{hc}{2e\lambda_{0}}$$V_{s}' = V_{0} - \frac{hc}{2e\lambda_{0}}$Note: If $\frac{hc}{2\lambda_{0}} < \phi$,the stopping potential will be $0$ as no photoelectric emission occurs.

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