Verified Solution • 2025–26 Session moderate

QNEET-2018-Ph-886: NEET 2018 Physics - Class 12 NEET-2018-Ph-886 Solution: An electron of mass m with an initial velocity \( {v} = v_0…

An electron of mass m with an initial velocity v⃗=v0i^\vec{v} = v_0\hat{i} (v0>0v_0 > 0) enters an electric field E⃗=−E0i^\vec{E} = -E_0\hat{i} (E0E_0 = constant > 0) at t = 0. If λ0\lambda_0 is its de-Broglie wavelength initially then its de-Broglie wavelength at time t is (1+eE0mv0t)λ0(1+\frac{eE_0}{mv_0}t)\lambda_0 λ0\lambda_0 (1+eE0mv0t)(1+\frac{eE_0}{mv_0}t) λ0t\lambda_0 t λ0t\frac{\lambda_0}{t}

Multiple Choice Options

  • A λ0(1+eE0mv0t){{{\lambda _0}} \over {\left( {1 + {{e{E_0}} \over {m{v_0}}}t} \right)}} ( 1 + m v 0 _ e E 0 _ _ t ) _ 0 _ _
  • B λ0(1+eE0mv0t){{\lambda _0}\left( {1 + {{e{E_0}} \over {m{v_0}}}t} \right)} _ 0 _ ( 1 + m v 0 _ e E 0 _ _ t )
  • C λ\lambda _ 0 t
  • D λ\lambda _ 0

Step-by-Step Verified Solution

Given, Initial velocity v = v 0 i ^ Electric field E = � E 0 i ^ _ Initial de-Brogile wavelength, _ 0 = h m v 0 Now force due to electric field on electrons, F _ = ( _ e ) ( _ E 0 i ^ ) = e E 0 i ^ Acceleration produced in the electron, a _ = F _ m = e E 0 m i ^ Now, velocity of electron after time t, v t _ = v _ + a _ t = ( v 0 + e E 0 t m ) i ^ _ | v t _ | = v 0 + e E 0 t m Now, _ t = h m v t = h m ( v 0 + e E 0 t m ) = h m v 0 ( 1 + e E 0 t m v 0 ) = _ 0 ( 1 + e E 0 t m v 0 ) [As _ 0 = h m v 0 ]

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