An electron (mass m) with initial velocity v = v _0i^{\^} + v_0j^{\^} is in an electric field E^{\to} = -E_ 0k^{\^}. If \lambda_0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by : [8 Jan 2020 Shift 2]

Correct Answer is : λ0√1+e2E2t22m2v02

{\lambda_0\sqrt2}/\sqrt{1+\frac{e^2E^2t^2}{m^2v_0^2}}

{\lambda_0 }/\sqrt{2+\frac{e^2E^2t^2}{m^2v_0^2}}

{\lambda_0 }/\sqrt{1+\frac{e^2E^2t^2}{2m^2v_0^2}}

{\lambda_0 }/\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}

JEE Main 8 Jan 2020 Shift 2 Solved Paper

Section: Physics

Question : 15 of 75

Marks: +1, -0

An electron (mass m) with initial velocity v=v0

+v0

is in an electric field

→

=−E0

. If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

[8 Jan 2020 Shift 2]

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