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  • The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum spe

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectivly will be :

Option: 1

1:1


Option: 2

2:1


Option: 3

4:1


Option: 4

1:4


Answers (1)

best_answer

 Work function =\mathrm{\phi_{0}=0.6 \mathrm{eV}}
\begin{array}{l} h f_{1}=3.8 \mathrm{eV} \\ h f_{2}=1.4 \mathrm{eV} \end{array}

From Einstein equation of photoelectric effect,
\begin{array}{l} h f=\phi_{0}+\frac{1}{2} m v_{\max }^{2} \\ \\ \therefore h f_{1}-\phi_{0}=\frac{1}{2} m v_{1}^{2} \rightarrow(1) \\ \\ h f_{2}-\phi_{0}=\frac{1}{2} m v_{2}^{2} \rightarrow(2) \\ \\ (1) \div(2) \\ \\ \frac{3 \cdot 2}{0 \cdot 8}=\frac{v_{1}^{2}}{v_{2}^{2}} \\ \\ \frac{v_{1}}{v_{2}}=2 \end{array}
The correct option is (2)

Posted by

rishi.raj

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