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The threshold frequency of a metal is \mathrm{f}_0 . When the light of frequency 2 \mathrm{f}_0 is incident on the metal plate, the maximum velocity of photoelectrons is v_1. When the frequency of incident radiation is increased to 5 \mathrm{f}_0, the maximum velocity of photoelectrons emitted is v_2. The ratio ofv_1$ to $v_2 is:

Option: 1

\frac{v_1}{v_2}=\frac{1}{8}


Option: 2

\frac{v_1}{v_2}=\frac{1}{4}


Option: 3

\frac{v_1}{v_2}=\frac{1}{16}


Option: 4

\frac{v_1}{v_2}=\frac{1}{2}


Answers (1)

best_answer

Using photoelectric equation
$$ \mathrm{hf}-\mathrm{hf}_0=\mathrm{eV}_0
As per question
$$ \begin{aligned} & \mathrm{h}\left(2 \mathrm{f}_0\right)-\mathrm{h}\left(\mathrm{f}_0\right)=\mathrm{eV}_1 \\ & \mathrm{~h}\left(2 \mathrm{f}_0-\mathrm{f}_0\right)=\mathrm{eV}_1 \\ & \mathrm{hf}_0=\mathrm{eV}_1 \, \, \, \, \, \, \, ...(1)\\ & \mathrm{~h}\left(5 \mathrm{f}_0\right)-\mathrm{hf}_0=\mathrm{eV}_2\\ & \mathrm{~h}\left(5 \mathrm{f}_0-\mathrm{f}_0\right)=\mathrm{eV}_2 \\ & 4 \mathrm{hf}_0=\mathrm{eV}_2 \, \, \, \, \, \, .... (2)\end{aligned}
Equation \frac{2}{1} \Rightarrow \frac{4 \mathrm{hf}_0}{\mathrm{hf}_0}=\frac{\mathrm{eV}_2}{\mathrm{eV}_1}
$$ \frac{\mathrm{V}_2}{\mathrm{~V}_1}=4
As we know
$$ \begin{aligned} & \mathrm{KE}_{\max }=\mathrm{eV}=\frac{1}{2} \mathrm{mv}_{\max }^2 \\ & \mathrm{v}_{\max } \propto \sqrt{\mathrm{V}} \\ & \therefore \frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{V}_2}{\mathrm{~V}_1}}=\sqrt{4}=2 \\ & \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1}{2} \end{aligned}

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jitender.kumar

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