JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

An electron of mass $m$ is moving in an electric field $E = - 2 E_{0} \hat{i}$ ( $E_{0}$ = constant $> 0$ ), with an initial velocity $V = v_{0} \hat{i}$ ( $v_{0}$ = constant $> 0$ ). If $λ_{0} = \frac{h}{4 m v _{0}}$ , its de Broglie wavelength at time $t$ is _________. ( $e$ = charge of electron)

A
$\frac{4 λ _{0}}{[ 1 - \frac{E _{0} e}{2 m} \frac{t}{v _{0}} ]}$
B
$\frac{4 λ _{0}}{[ 1 + \frac{E _{0} e}{2 m} \frac{t}{v _{0}} ]}$
C
$\frac{4 λ _{0}}{[ 1 + \frac{2 E _{0} e}{m} \frac{t}{v _{0}} ]}$
D
$\frac{4 λ _{0}}{[ 1 - \frac{2 E _{0} e}{m} \frac{t}{v _{0}} ]}$

Solution

Force on the electron: $F = (- e) E = (- e) (- 2 E_{0} \hat{i}) = 2 e E_{0} \hat{i}$ , so its acceleration is $a = \frac{2 e E _{0}}{m}$ along $+ \hat{i}$ , same direction as the initial velocity.

Velocity at time $t$ : $v (t) = v_{0} + \frac{2 e E _{0}}{m} t = v_{0} (1 + \frac{2 e E _{0} t}{m v _{0}})$ .

De Broglie wavelength: $λ = \frac{h}{m v ( t )}$ . Using $h = 4 m v_{0} λ_{0}$ :

$λ = \frac{4 m v _{0} λ _{0}}{m v _{0} ( 1 + \frac{2 e E _{0} t}{m v _{0}} )} = \frac{4 λ _{0}}{1 + \frac{2 E _{0} e}{m} \frac{t}{v _{0}}}$ .

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