Q185 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
The de Broglie wavelength of an electron with kinetic energy $K_1$ is $\lambda_1$. If the kinetic energy increases to $5K_1$, the new wavelength is $\lambda_2 = x\lambda_1$. The value of $x$ is
- A$5$
- B$\sqrt{5}$
- Correct answerC$\dfrac{1}{\sqrt{5}}$
- D$\dfrac{1}{5}$
Solution
The de Broglie wavelength of an electron is given by $\lambda = \dfrac{h}{\sqrt{2mK}}$
For the initial state, $\lambda_1 = \dfrac{h}{\sqrt{2mK_1}}$
For the final state, the kinetic energy is $K_2 = 5K_1$. The new wavelength is:
$\lambda_2 = \dfrac{h}{\sqrt{2m(5K_1)}} = \dfrac{1}{\sqrt{5}} \dfrac{h}{\sqrt{2mK_1}}$
$\lambda_2 = \dfrac{1}{\sqrt{5}} \lambda_1$
Comparing this with $\lambda_2 = x\lambda_1$, we get $x = \dfrac{1}{\sqrt{5}}$
Answer: $\dfrac{1}{\sqrt{5}}$
For the initial state, $\lambda_1 = \dfrac{h}{\sqrt{2mK_1}}$
For the final state, the kinetic energy is $K_2 = 5K_1$. The new wavelength is:
$\lambda_2 = \dfrac{h}{\sqrt{2m(5K_1)}} = \dfrac{1}{\sqrt{5}} \dfrac{h}{\sqrt{2mK_1}}$
$\lambda_2 = \dfrac{1}{\sqrt{5}} \lambda_1$
Comparing this with $\lambda_2 = x\lambda_1$, we get $x = \dfrac{1}{\sqrt{5}}$
Answer: $\dfrac{1}{\sqrt{5}}$
Hence, the correct answer is option: 'C'.