Q161 / 180 · COMEDK 2025 (25 May Shift) · Physics
The de-Broglie wavelength of a neutron having kinetic energy 0.025 eV is $\lambda$. If the kinetic energy of the neutron is reduced to $\dfrac{0.025 \mathrm{eV}}{8}$, its de-Broglie wavelength will be:
- Correct answerA$2 \sqrt{2} \lambda$
- B$\dfrac{\lambda}{2 \sqrt{2}}$
- C$\dfrac{0.025 \lambda}{2 \sqrt{2}}$
- D$\dfrac{\lambda}{0.025 \sqrt{2}}$
Solution
The de-Broglie wavelength $\lambda$ of a particle with mass $m$ and kinetic energy $K$ is given by the relation $\lambda = \dfrac{h}{\sqrt{2mK}}$.
For the initial state, the wavelength is $\lambda = \dfrac{h}{\sqrt{2m(0.025 \text{ eV})}}$.
When the kinetic energy is reduced to $K' = \dfrac{0.025 \text{ eV}}{8}$, the new wavelength $\lambda'$ is given by $\lambda' = \dfrac{h}{\sqrt{2m \times \dfrac{0.025 \text{ eV}}{8}}}$.
Simplifying the expression for $\lambda'$:
$\lambda' = \dfrac{h}{\sqrt{\dfrac{2m(0.025 \text{ eV})}{8}}} = \sqrt{8} \times \dfrac{h}{\sqrt{2m(0.025 \text{ eV})}}$.
Since $\sqrt{8} = 2\sqrt{2}$, we have $\lambda' = 2\sqrt{2} \lambda$.
Answer: $2 \sqrt{2} \lambda$
For the initial state, the wavelength is $\lambda = \dfrac{h}{\sqrt{2m(0.025 \text{ eV})}}$.
When the kinetic energy is reduced to $K' = \dfrac{0.025 \text{ eV}}{8}$, the new wavelength $\lambda'$ is given by $\lambda' = \dfrac{h}{\sqrt{2m \times \dfrac{0.025 \text{ eV}}{8}}}$.
Simplifying the expression for $\lambda'$:
$\lambda' = \dfrac{h}{\sqrt{\dfrac{2m(0.025 \text{ eV})}{8}}} = \sqrt{8} \times \dfrac{h}{\sqrt{2m(0.025 \text{ eV})}}$.
Since $\sqrt{8} = 2\sqrt{2}$, we have $\lambda' = 2\sqrt{2} \lambda$.
Answer: $2 \sqrt{2} \lambda$
Hence, the correct answer is option: 'A'.