An electron of mass m with an initial velocity → V = V _ 0 ^ i ( V _ 0 > 0 ) enters in an electric field → E = -E _ 0 ^ i ( E _ 0 > 0 and is a constant ) at t = 0 . If λ _ 0 is its de-Broglie wavelength initially, then

Correct answer: C. λ _ 0 (1 + e E _ 0 m V _ 0 t) → V = V _ 0 ^ i → E = - E _ 0 ^ i → a = → F m = q → E m = (- e)(- E _ 0 ) ^ i m → a = e E _ 0 m ^ i V = u + at V = V _ 0 + e E _ 0 m t ∵ De Broglie wavelength= λ = (h)/(mv) and initially λ _ 0 = h m v _ 0 . = h m[ V _ 0 + e E _ 0 t m ] = h m V _ 0 [1 + e E _ 0 t m V _ 0 ] = λ _ 0 1 + e E _ 0 t m V _ 0

NEET 2018

Exam: NEET
Year: 2018
Subject: Physics
Chapter: Dual Nature of Matter

NEET 2018 · Physics · Dual Nature of Matter

An electron of mass m with an initial velocity → V = V _ 0 ^ i ( V...

An electron of mass

m

with an initial velocity

\overset{\to}{V} = {V}_{0}\overset{^}{i}( {V}_{0} > 0 )

enters in an electric field

\overset{\to}{E} = {-E}_{0}\overset{^}{i}( {E}_{0} > 0

and is a constant

)

t = 0

. If

{\lambda}_{0}

is its de-Broglie wavelength initially, then its de-Broglie wavelength at time

t

An electron of mass m with an initial velocity → V = V _ 0 ^ i ( V _ 0 > 0 ) enters in an electric field → E = -E _ 0 ^

Options

A ${\lambda}_{0}t$
B ${\lambda}_{0}\left(1 + \dfrac{e{E}_{0}}{m{V}_{0}}t\right)$
C\dfrac{{\lambda}_{0}}{ $\left(1 + \dfrac{e{E}_{0}}{m{V}_{0}}t\right)}$
D ${\lambda}_{0}$

Correct answer

C. \dfrac{{\lambda}_{0}}{ $\left(1 + \dfrac{e{E}_{0}}{m{V}_{0}}t\right)}$

Step-by-step solution

\overset{\to}{V} = {V}_{0}\overset{^}{i}

\overset{\to}{E} = - {E}_{0}\overset{^}{i}

\overset{\to}{a} = \overset{\to}{F}m = \dfrac{q\overset{\to}{E}}{m} = \dfrac{\left(- e\right)\left(- {E}_{0}\right)\overset{^}{i}}{m}

\overset{\to}{a} = \dfrac{e{E}_{0}}{m}\overset{^}{i}

V = u + at

V = {V}_{0} + \dfrac{e{E}_{0}}{m}t

∵

De Broglie wavelength=

\lambda = \dfrac{h}{mv}

and initially

{\lambda}_{0} = \dfrac{h}{m{v}_{0}}

= \dfrac{h}{m\left[{V}_{0} + \dfrac{e{E}_{0}t}{m}\right]} = \dfrac{h}{m{V}_{0}\left[1 + \dfrac{e{E}_{0}t}{m{V}_{0}}\right]} = \dfrac{{\lambda}_{0}}{1 + \dfrac{e{E}_{0}t}{m{V}_{0}}}

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