MHT CET · Physics · Dual Nature of Matter
If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
- A 75
- B 25
- C 50
- D 5
Answer & Solution
Correct Answer
(A) 75
Step-by-step Solution
Detailed explanation
Kinetic energy \(k=\frac{p^{2}}{2 m}\) where \(p\) is momentum
\(
\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{P}_{2}^{2}}{\mathrm{P}_{1}^{2}} \quad \therefore \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}=\sqrt{16}=4
\)
de Broglie wavelength \(\lambda=\frac{\mathrm{h}}{\mathrm{P}}\)
\(
\therefore \frac{\lambda_{2}}{\lambda_{1}}=\frac{P_{1}}{P_{2}}=\frac{1}{4} \quad \therefore \lambda_{2}=\frac{\lambda_{1}}{4} \quad \therefore \lambda_{1}-\lambda_{2}=\frac{3}{4} \lambda_{1}
\)
\(
\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{P}_{2}^{2}}{\mathrm{P}_{1}^{2}} \quad \therefore \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}=\sqrt{16}=4
\)
de Broglie wavelength \(\lambda=\frac{\mathrm{h}}{\mathrm{P}}\)
\(
\therefore \frac{\lambda_{2}}{\lambda_{1}}=\frac{P_{1}}{P_{2}}=\frac{1}{4} \quad \therefore \lambda_{2}=\frac{\lambda_{1}}{4} \quad \therefore \lambda_{1}-\lambda_{2}=\frac{3}{4} \lambda_{1}
\)
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