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  • An electron (mass m) with an initial velocity v = v0i(v0 > 0) is in an electric field E=E_{0}hat{i} (E0 = constant > 0). Its de-Broglie wavelength at time t is given by(a)frac{lambda_{0}}{[ 1+frac{eE_{0}}{m}frac{t}{v_{0}}]}

An electron (mass m) with an initial velocity v = v0i(v0 > 0) is in an electric field E=E_{0}\hat{i} (E0 = constant > 0). Its de-Broglie wavelength at time t is given by

(a) \frac{\lambda_{0}}{\left [ 1+\frac{eE_{0}}{m}\frac{t}{v_{0}} \right ]}\\ (b)\lambda_{0}\left [ 1+\frac{eE_{0}t}{mv_{0}} \right ]\\ (c)\lambda_{0}\\ (d)\lambda_{0}t

Answers (1)

The answer is the option (a)
Initial de-Broglie wavelength \lambdais given by

\lambda_{0}=\frac{h}{p}=\frac{h}{mv_{0}}

  Force on electron=F=qE

ma=eE_{0}\hat{i}\\ a=\frac{eE}{m}\hat{i}

   Velocity of electron after time t is v=v0+at 

 v=v_{0}i+\frac{eE}{m}i.t\\ v=\left [ v_{0}+\frac{eEt}{m} \right ]\hat{i}\\

 New de-Broglie wavelength

\lambda=\frac{h}{mv}

\lambda=\frac{h}{m\left [ v_{0}+\frac{eE_{0}t}{m} \right ]}\\ \lambda=\frac{\lambda_{0}}{\left [ 1+\frac{eE_{0}t}{mv_{0}} \right ]}\; \; \; \left ( Since \frac{h}{mv_{0}}=\lambda_{0} \right )

    

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