Q2 / 160 · TS EAMCET 2023 (12 May Shift 1) · Chemistry

The threshold frequency of a metal is $10^{15} \mathrm{~s}^{-1}$. The ratio of maximum kinetic energies of the photoelectrons, when the metal is made to strike with radiations of frequencies $1.5 \times 10^{15} \mathrm{~s}^{-1}$ and $2.0 \times 10^{15} \mathrm{~s}^{-1}$ respectively is

  • A
    $2:1$
  • Correct answer
    B
    $1:2$
  • C
    $4:3$
  • D
    $3:4$

Solution

$v_0=1.0 \times 10^{15} \mathrm{~s}^{-1}$ Maximum kinetic energy will be at the highest frequencies and when all the energy is utilized for the kinetic energy of the ejected electrons: $\begin{aligned} & \Rightarrow \mathrm{K} . \mathrm{E}_1=\mathrm{h}\left(v_1-v_0\right) \\ & =6.626 \times 10^{-34}\left(1.5 \times 10^{15}-1.0 \times 10^{15}\right) \\ & =6.626 \times 10^{-34}\left(0.5 \times 10^{15}\right) \\ & =3.313 \times 10^{-19} \mathrm{~J} \\ & \Rightarrow \mathrm{K}_{.} \mathrm{E}_2=\mathrm{h}\left(v_2-v_0\right) \\ & =6.626 \times 10^{-34}\left(2.0 \times 10^{15}-1.0 \times 10^{15}\right) \\ & =6.626 \times 10^{-34}\left(1 \times 10^{15}\right) \\ & =6.626 \times 10^{-19} \mathrm{~J} \end{aligned}$ Thus, the ratio will be : $\frac{\mathrm{K} . \mathrm{E}_1}{\mathrm{~K} . \mathrm{E}_2}=\frac{3.313 \times 10^{-19}}{6.626 \times 10^{-19}}=1: 2$

Hence, the correct answer is option: 'B'.

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