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

(D) $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} E^{2} t^{2}}{2 m^{2} v_{0}^{2}}}}$

(A) $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

(B) $\frac{\lambda_{0}}{\sqrt{2+\frac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

(C) $\frac{\lambda_{0} \sqrt{2}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}$

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Modern Physics

An electron (mass

m

) with initial velocity

v = v_{0} \hat{i} + v_{0} \hat{j}

is in an electric field

E = - E_{0} \hat{k}

. If

λ_{0}

is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time

t

is given by

(A) $\frac{λ _{0}}{1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{m ^{2} v _{0}^{2}}}$
(B) $\frac{λ _{0}}{2 + \frac{e ^{2} E ^{2} t ^{2}}{m ^{2} v _{0}^{2}}}$
(C) $\frac{λ _{0} 2}{1 + \frac{e ^{2} E ^{2} t ^{2}}{m ^{2} v _{0}^{2}}}$
(D) $\frac{λ _{0}}{1 + \frac{e ^{2} E ^{2} t ^{2}}{2 m ^{2} v _{0}^{2}}}$ ✓ Correct

Answer: (D)

The initial velocity

v = v_{0} \hat{i} + v_{0} \hat{j}

has magnitude

2 v_{0}

, so the initial momentum is

m 2 v_{0}

and

λ_{0} = \frac{h}{m 2 v _{0}}

. The field

E = - E_{0} \hat{k}

pushes the electron with force

e E_{0}

along

\hat{k}

, giving it a growing

z

-component of velocity

\frac{e E _{0}}{m} t

while the

x

and

y

components are untouched. At time

t

the speed satisfies

v^{2} = 2 v_{0}^{2} + (\frac{e E _{0} t}{m})^{2}

, so the de-Broglie wavelength is

λ = \frac{h}{m 2 v _{0}^{2} + e ^{2} E _{0}^{2} t ^{2} / m ^{2}}

. Dividing by

λ_{0}

and taking

2 v_{0}^{2}

out of the root leaves

λ = \frac{λ _{0}}{1 + \frac{e ^{2} E _{0}^{2} t ^{2}}{2 m ^{2} v _{0}^{2}}}

, where the factor of two in the denominator comes straight from the two equal initial velocity components.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option