Atomic Structure Result Question 6
6. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $h / \lambda$ (where, $\lambda$ is wavelength associated with electron wave) is given by
(2016 Main)
(a) $2$ meV
(b) $\sqrt{\mathrm{meV}}$
(c) $\sqrt{2 m e V}$
(d) meV
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Answer:
Correct Answer: 6. ( c )
Solution:
- Plan: As you can see in options, energy term is mentioned hence, we have to find out relation between $h / \lambda$ and energy. For this, we shall use de-Broglie wavelength and kinetic energy term in eV .
de-Broglie wavelength for an electron $(\lambda)=\dfrac{h}{p}$
$\Rightarrow \quad p=\dfrac{h}{\lambda}\quad $ ……(i)
Kinetic energy of an electron $=\mathrm{eV}$
As we know that, $\mathrm{KE}=\dfrac{p^2}{2 m}$
$\therefore \quad \mathrm{eV}=\dfrac{p^2}{2 m} \quad$ or $\quad p=\sqrt{2 m e V}\quad $ ……(ii)
From equations (i) and (ii), we get $\dfrac{h}{\lambda}=\sqrt{2 \mathrm{meV}}$
BETA
