MHT CET · Physics · Dual Nature of Matter
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 \(4.14 \times 10^3 \mathrm{MeV}\)
- D \(4.14 \times 10^3 \mathrm{eV}\)
Answer & Solution
Correct Answer
(D) \(4.14 \times 10^3 \mathrm{eV}\)
Step-by-step Solution
Detailed explanation
Using plank energy quanta vs frequency relation:
\(E(e V)=\frac{h v}{e}=\frac{6.63 \times 10^{-34} \times 10^{12} \times 10^6}{1.6 \times 10^{-19}}=4.14 \times 10^3 \mathrm{eV}\)
\(E(e V)=\frac{h v}{e}=\frac{6.63 \times 10^{-34} \times 10^{12} \times 10^6}{1.6 \times 10^{-19}}=4.14 \times 10^3 \mathrm{eV}\)
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