Q156 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Physics

A metal has work function $2.5 \mathrm{eV}$. If a radiation of frequency $32 \times 10^{15} \mathrm{~Hz}$ is incident on this metal surface, then the maximum kinetic energy of ejected photoelectrons is (Planck's constant, $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

  • A
    $9.5 \mathrm{eV}$
  • B
    $2.5 \mathrm{eV}$
  • Correct answer
    C
    $10.7 \mathrm{eV}$
  • D
    $12.6 \mathrm{eV}$

Solution

Frequency of incident radiation, $f=3.2 \times 10^{15} \mathrm{~Hz}$ Energy of incident photon $E, E=h f$ $\begin{aligned} & =6.6 \times 10^{-34} \times 3.2 \times 10^{15} \\ & =\frac{6.6 \times 10^{-34} \times 3.2 \times 10^{15}}{1.6 \times 10^{-19}} \mathrm{eV} \\ & =13.2 \mathrm{eV}\end{aligned}$ So, maximum K. E of emitted photoelectrons $=E-\phi_0$ $=13.2-25$ $\left(\because \phi_0=2.5 \mathrm{eV}\right.$, given $)$ $=10.7 \mathrm{eV}$

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

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