Question

An electron (mass m) with an initial velocity `v = v_0hati` is in an electric field `E = E_0hatj`. If λ₀ = h/mv₀, it’s de Broglie wavelength at time t is given by ______.

पर्याय

• `λ_0`

• `λ_0 sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2))`

• `λ_0/sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2))`

• `λ_0/((1 + (e^2E_0^2t^2)/(m^2v_0^2))`

Answer 1

An electron (mass m) with an initial velocity `v = v_0hati` is in an electric field `E = E_0hatj`. If λ₀ = h/mv₀, it’s de Broglie wavelength at time t is given by `underline(λ_0/sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2)))`.

Explanation:

According to the problem de-Broglie wavelength of electron at time t = 0 is `λ_0 = h/(mv_0)`

Electrostatic force on electron in electric field is `vecF_e = - evecE = - eE_0hatj`

The acceleration of electron, `veca = vecF/m = - (eE_0)/m hatj`

It is acting along a negative y-axis.

The initial velocity of electron along x-axism `v_(x_0) = v_0hati`.

This component of velocity will remain constant as there is no force on the electron in this direction.

Now considering y-direction. Initial velocity of electron along y-axis, `v_(y_0)` = 0.

Velocity of electron after time t along y-axis,

`v_y = 0 + ((eE_0)/m hatj)t = - (eE_0)/m t hatj`

Magnitude of velocity of electron after time t is

`v = sqrt(v_x^2 + v_y^2) = sqrt(v_0^2 + ((-eE_0)/m t)^2`

⇒ `v_0 sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2))`

de-Broglie wavelength, `λ^' = h/(mv)`

= `h/(mv_0 sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2))) = λ_0/sqrt(1 + (e^2E_0^2t^2)/(m^2v_0^2))`

⇒ `λ^' = λ_0/((1 + (e^2E_0^2t^2)/(m^2v_0^2))`