MHT CET · Physics · Dual Nature of Matter
How much energy is imparted to an electron so that its de-Broglie wavelength
reduces from \(10^{-10} \mathrm{~m}\) to \(0 \cdot 5 \times 10^{-10} \mathrm{~m} ?(\mathrm{E}=\) energy of electron \()\)
- A \(4 \mathrm{E}\)
- B \(2 \mathrm{E}\)
- C \(3 \mathrm{E}\)
- D \(\mathrm{E}\)
Answer & Solution
Correct Answer
(C) \(3 \mathrm{E}\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\)
\(\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}} \Rightarrow \frac{1 \times 10^{-9}}{0.5 \times 10^{-9}}=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}}\)
\(\therefore 2=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}} \quad \therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=4\)
\(\therefore \mathrm{E}_{2}=4 \mathrm{E}_{1}\)
\(\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}} \Rightarrow \frac{1 \times 10^{-9}}{0.5 \times 10^{-9}}=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}}\)
\(\therefore 2=\sqrt{\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}} \quad \therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=4\)
\(\therefore \mathrm{E}_{2}=4 \mathrm{E}_{1}\)
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