A charge particle q-0 of mass m-0 is projected along the y-axis t - 0 from origin with a velocity V-0- If a uniform electric field E-0 also exists along the x-axis- then the time which de Broglie wavelength of the particle becomes half of the initial value is -dfrac-m-0-v-0-q-0-E-0-2dfrac-m-0-v-0-q-0-E-0-sqrt3dfrac-m-0-v-0-q-0-E-0-3dfrac-m-0-v-0-q-0-E-0-

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Toppr Admin · Accepted Answer

Initial debrogle wavelength of the charge particle $λ_{0} = \frac{h}{m_{0} v_{0}}$
After any time t, $λ^{'} = \frac{h}{m_{0} \sqrt{V_{0}^{2} + {(\frac{q E_{0} t}{m_{0}})}^{2}}}$
Suppose after time t, $λ$ becomes half of the initial value i.e. $λ = λ_{0} / 2$ .
$t = \sqrt{3} \frac{m_{0} v_{0}}{q_{0} E_{0}}$

A charge particle q0 of mass m0 is projected along the y-axis at t=0 from origin with a velocity V0. If a uniform electric field E0 also exists along the x-axis, then the time at which de Broglie wavelength of the particle becomes half of the initial value is :m0v0q0E02m0v0q0E0√3m0v0q0E03m0v0q0E0

Initial debrogle wavelength of the charge particle λ0=hm0v0After any time t, λ′=hm0√V20+(qE0tm0)2Suppose after time t, λ becomes half of the initial value i.e. λ=λ0/2.t=√3m0v0q0E0