MHTCET 25 Apr 2025 Shift 2 Paper

Concept:A charged particle in a uniform electric field experiences constant force, producing uniform acceleration. As the electron gains velocity, its de-Broglie wavelength changes with time, and the required rate is obtained by differentiating $\lambda$ with respect to $t$.Explanation:The electric force acting on the electron is $F = eE$.Using Newton's second law, the acceleration is $a = \frac{F}{m} = \frac{eE}{m}$.The electron is initially at rest, so its initial velocity is $u = 0$.After time $t$, the velocity acquired is $v = u + at = 0 + \frac{eE}{m} \cdot t = \frac{eEt}{m}$.The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Substituting the expression for $v$, we get $\lambda = \frac{h}{m \cdot \frac{eEt}{m}} = \frac{h}{eEt}$.To find the rate of change of wavelength, differentiate $\lambda$ with respect to $t$:$\frac{d\lambda}{dt} = \frac{d}{dt}\left(\frac{h}{eEt}\right) = \frac{h}{eE} \cdot \frac{d}{dt}(t^{-1}) = \frac{h}{eE} \cdot \left(-\frac{1}{t^2}\right)$.Therefore, $\frac{d\lambda}{dt} = -\frac{h}{eEt^2}$.The negative sign indicates that the de-Broglie wavelength decreases as the electron accelerates.Answer:The rate of change of de-Broglie wavelength is $-\frac{h}{eEt^2}$, which matches option A.