WBJEE · Physics · Dual Nature of Matter
The de-Broglie wavelength of an electron is \(0.4 \times 10^{-10} \mathrm{m}\) when its kinetic energy is \(1.0 \mathrm{keV} .\) Its wavelength will be \(1.0 \times 10^{-10} \mathrm{m}\)
when its kinetic energy is
- A \(0.2 \mathrm{keV}\)
- B \(0.8 \mathrm{keV}\)
- C \(0.63 \mathrm{keV}\)
- D \(0.16 \mathrm{keV}\)
Answer & Solution
Correct Answer
(D) \(0.16 \mathrm{keV}\)
Step-by-step Solution
Detailed explanation
De-Broglie wavelength, \(\lambda=\frac{h}{\sqrt{2 m E}}\) \(\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{2}}{E_{1}}}\) \(\lambda_{1}=\) Wave length of electron…
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