JEE PYQ: Atomic Physics Question 41
Question 41 - 2020 (08 Jan Shift 2)
An electron (mass m) with initial velocity $\vec{v} = v_0\hat{i} + v_0\hat{j}$ is in an electric field $\vec{E} = -E_0\hat{k}$. If $\lambda_0$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time $t$ is given by:
(1) $\frac{\lambda_0\sqrt{2}}{\sqrt{1 + \frac{e^2E^2t^2}{m^2v_0^2}}}$
(2) $\frac{\lambda_0}{\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}}}$
(3) $\frac{\lambda_0}{\sqrt{1 + \frac{e^2E^2t^2}{2m^2v_0^2}}}$
(4) $\frac{\lambda_0}{\sqrt{2 + \frac{e^2E^2t^2}{m^2v_0^2}}}$
Show Answer
Answer: (3)
Solution
Initial velocity $u = v_0\hat{i} + v_0\hat{j}$. Acceleration $a = \frac{eE_0}{m}$. $v = v_0\hat{i} + v_0\hat{j} + \frac{eE_0}{m}t\hat{k}$. $|v| = \sqrt{2v_0^2 + \left(\frac{eE_0t}{m}\right)^2}$. $\lambda_0 = \frac{h}{mv_0\sqrt{2}}$. $\lambda = \frac{h}{m\sqrt{2v_0^2 + \frac{e^2E_0^2t^2}{m^2}}}$. $\frac{\lambda}{\lambda_0} = \frac{1}{\sqrt{1 + \frac{e^2E_0^2t^2}{2m^2v_0^2}}}$. So $\lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2E^2t^2}{2m^2v_0^2}}}$.
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