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प्रश्न
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 ______.
विकल्प
`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))`
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उत्तर
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))`
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