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An electron of mass \mathrm{m} and charge e initially at rest gets accelerated by a constant electric field \mathrm{E}. The rate of change of de-Broglie wavelength of this electron at time \mathrm{t} (ignoring relativistic effects) is:
 

Option: 1

\frac{-\mathrm{h}}{\mathrm{eE} \mathrm{t}^2}


Option: 2

\frac{-\mathrm{mh}}{\mathrm{eEt}^2}


Option: 3

\frac{-\mathrm{h}}{\mathrm{eEt}}


Option: 4

\frac{-\mathrm{eEt}}{\mathrm{h}}


Answers (1)

best_answer


Here, \mathrm{u}=0, \mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}

 \therefore \mathrm{v}=\mathrm{u}+\mathrm{at}=0+\frac{\mathrm{eE}}{\mathrm{m}} \mathrm{t}

de - Broglie wavelength, 

\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{eEt} / \mathrm{m})}=\frac{\mathrm{h}}{\mathrm{eEt}}

Rate of change of de-Broglie wavelength

 \frac{\mathrm{d} \lambda}{\mathrm{dt}}=\frac{\mathrm{h}}{\mathrm{eE}}\left(-\frac{1}{\mathrm{t}^2}\right)=\frac{-\mathrm{h}}{\mathrm{eEt}}

Posted by

Ajit Kumar Dubey

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