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  • Find the frequency of light which eject electrons from a metal surface fully stopped by a retarding potential of  <img alt="3 \mathrm{~V}" src="https://entrancecorner.oncodecogs.com/gif.l

Find the frequency of light which eject electrons from a metal surface fully stopped by a retarding potential of  3 \mathrm{~V}.The photo electric effect begins in this metal at the frequency of  6 \times 10^{14} \ s^{-1}.

Option: 1

1.1 \times 10^{15} \mathrm{~Hz}


Option: 2

2.1 \times 10^{15} \mathrm{~Hz}


Option: 3

1.32 \times 10^{15} \mathrm{~Hz}


Option: 4

1.45 \times 10^{15} \mathrm{~Hz}


Answers (1)

best_answer

The threshold frequency for the given metal surface is  \nu_{\text {th }}=6 \times 10^{14} \mathrm{~Hz}.

Thus the work function for metal surface is
\begin{aligned} \phi & =h \nu_{\text {th }}=6.63 \times 10^{-34} \times 6 \times 10^{14} \\ & =3.98 \times 10^{-19} \mathrm{~J} . \end{aligned}

As stopping potential for the ejected electrons is 3 \mathrm{~V}, the maximum kinetic energy of ejected electrons will be
K \cdot E_{\text {max }}=3 \mathrm{eV}=3 \times 1.6 \times 10^{-19} \mathrm{~J}=4.8 \times 10^{-19} \mathrm{~J} \text {. }
According to photo electric effect equation  we have


h \nu=h \nu_{th}+K \cdot E_{\max }=\phi+K \cdot E_{\text {max }}
\RightarrowIncident frequency is


$$ \begin{aligned} \nu & =\frac{\phi+K \cdot E_{\text {max }}}{h}=\frac{3.98 \times 10^{-19}+4.8 \times 10^{-19}}{6.63 \times 10^{-34}} \\ & =1.32 \times 10^{15} \mathrm{~Hz} . \end{aligned}

Posted by

Rakesh

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