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When two monochromatic lights of frequency, \nu and \frac{\nu}{2} are incident on a photoelectric metal, their stopping potential becomes \mathrm{\frac{Vs}{2}} and \mathrm{Vs} respectively. The threshold frequency for this metal is :

Option: 1

3 \nu


Option: 2

\frac{2}{3} \nu


Option: 3

\frac{3}{2} \nu


Option: 4

2 \nu


Answers (1)

best_answer

By Einstein's photoelectric equation,

for the first source

\mathrm{h \nu=\phi_{0}+\frac{e V_{s}}{2}}      .......(1)

For the second source,

\mathrm{\frac{h \nu}{2}=\phi_{0}+e V_{s}}     ..........(2)

2\times Eqn (1) - Eqn(2)

\mathrm{\frac{3 h \nu}{2}=\phi_{0}=h \nu_{0}}\\

\mathrm{\nu_{0} \rightarrow threshold\; frequency}\\

\mathrm{\nu_{0} \equiv \frac{3\nu}{2}}

hence correct option is 3

Posted by

Gaurav

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