Q122 / 150 · MHT CET 2022 (07 Aug Shift 2) · Physics

Energy of photon whose frequency is $10^{12} \mathrm{MHz}$ is [Planck's constant, $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ]

  • A
    $4.14 \times 10^3 \mathrm{keV}$
  • B
    $4.14 \times 10^2 \mathrm{eV}$
  • C
    $4.14 \times 10^3 \mathrm{MeV}$
  • Correct answer
    D
    $4.14 \times 10^3 \mathrm{eV}$

Solution

Using plank energy quanta vs frequency relation: $E(e V)=\frac{h v}{e}=\frac{6.63 \times 10^{-34} \times 10^{12} \times 10^6}{1.6 \times 10^{-19}}=4.14 \times 10^3 \mathrm{eV}$

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

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