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QJEEM23Ph14e9: JEE Main 2023 Physics - Class 12 JEEM23Ph14e9 Solution: The threshold frequency of a metal is . When the light of…

The threshold frequency of a metal is f0f_{0} . When the light of frequency 2f02 f_{0} is incident on the metal plate, the maximum velocity of photoelectrons is v1v_{1} . When the frequency of incident radiation is increased to 5f05 \mathrm{f}_{0} , the maximum velocity of photoelectrons emitted is v2v_{2} . The ratio of v1v_{1} to v2v_{2} is :

Multiple Choice Options

  • A v1v2=12\frac{v_{1}}{v_{2}}=\frac{1}{2}
  • B v1v2=116\frac{v_{1}}{v_{2}}=\frac{1}{16}
  • C v1v2=14\frac{v_{1}}{v_{2}}=\frac{1}{4}
  • D v1v2=18\frac{v_{1}}{v_{2}}=\frac{1}{8}

Step-by-Step Verified Solution

amp;12mv2=hf−hf0amp;⇒12mv12=2hf0−hf0=hf0amp; also, 12mv22=5hf0−hf0=4hf0\begin{aligned}& \frac{1}{2} m v^2=h f-h f_0 \\\\& \Rightarrow \frac{1}{2} m v_1^2=2 h f_0-h f_0=h f_0 \\\\& \text { also, } \frac{1}{2} m v_2^2=5 h f_0-h f_0=4 h f_0\end{aligned}

taking ratio,

v12v22=14⇒v1v2=12\frac{v_1^2}{v_2^2}=\frac{1}{4} \Rightarrow \frac{v_1}{v_2}=\frac{1}{2}

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