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Question

Physics Question on de broglie hypothesis

The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100 V100 \,V is

A

9.1×10−249.1 \times10^{-24}

B

6.5×10−246.5 \times10^{-24}

C

5.4×10−245.4 \times10^{-24}

D

1.6×10−241.6 \times10^{-24}

Answer

5.4×10−245.4 \times10^{-24}

Explanation

Solution

de-Broglie wavelength of electron is given by
λ=hmv=h2mE\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}
Substituting the value of EE, we get
λ=h2meV\lambda=\frac{h}{\sqrt{2meV}}
Here m=9.1×10−31kg;e=1.6×10−19Cm=9.1\times10^{-31} kg; e=1.6\times10^{-19}C
and h=6.6×10−34Jsh=6.6\times10^{-34}Js
we get λ=12.27V×10−10=12.27V?\lambda=\frac{12.27}{\sqrt{V}}\times10^{-10}=\frac{12.27}{\sqrt{V}}?
The de-Broglie wavelength of electrons, when accelerated through a potential difference of 100 V100\, V will be
λ=12.27100=1.227?\lambda=\frac{12.27}{\sqrt{100}}=1.227?
Moreover, λ=hp\lambda=\frac{h}{p}
⇒p=6.6×10−341.227×10−10\Rightarrow p=\frac{6.6\times10^{-34}}{1.227\times10^{-10}}
=5.5×10−24kg−ms−1=5.5\times10^{-24} kg -ms^{-1}