WBJEE · Physics · Electromagnetic Waves
The energy of gamma \((\gamma)\) ray photon is \(E\), and that of an X-ray photon is \(E_{X}\). If the visible light photon has an energy of \(E_{v}\), then we can say that
- A \(E_{x}>E_{y}>E_{v}\)
- B \(E_{\gamma}>E_{y}>E_{x}\)
- C \(E_{\gamma}>E_{x}>E_{v}\)
- D \(E_{x}>E_{\gamma}>E_{y}\)
Answer & Solution
Correct Answer
(C) \(E_{\gamma}>E_{x}>E_{v}\)
Step-by-step Solution
Detailed explanation
\(E_{y} \geq 100 \mathrm{keV}\) \(\begin{aligned} E_{X} &=100 \mathrm{eV} \text { to } 100 \mathrm{keV} \\ E_{v} &=2.48 \mathrm{eV} \end{aligned}\) So, we can say that \(E_{y}>E_{X}>E_{v}\)
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