An electron (mass m) with an initial velocity $\vec{v} = v_{0} \hat{i} (v_{0} > 0)$ is moving in an electric field $\vec{E} = - E_{0} \hat{i} (E_{0} > 0)$ where E₀ is constant. If at t = 0 de Broglie wavelength is $λ_{0} = \frac{h}{m v_{0}}$ , then its de Broglie wavelength after time t is given by

Option 1 -

$λ_{0}$

Option 2 -

$λ_{0} (1 + \frac{e E_{0} t}{m v_{0}})$

Option 3 -

$λ_{0} t$

Option 4 -

$\frac{λ_{0}}{1 + \frac{e E_{0} t}{m v_{0}}}$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Given that $\vec{v} = v_{0} i a n d \vec{E} = - E_{0} \hat{i}$

$\vec{v} (t) = v_{0} \hat{i} + \frac{e E t}{m} \hat{i}$

So, de Broglie wave length, $λ = \frac{h}{m v}$

$\Rightarrow λ = \frac{h}{m v_{0} (1 + \frac{q E_{0}}{m v_{0}} t)} = \frac{λ_{0}}{(1 + \frac{q E_{0}}{m v_{0}} t)}$

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An electron (mass m) with an initial velocity v → = v 0 i ^ ( v 0 > 0 ) is moving in an electric field E → = − E 0 i ^ ( E 0 > 0 ) where E0 is constant. If at t = 0 de Broglie wavelength is λ 0 = h m v 0 , then its de Broglie wavelength after time t is given by