A stream of electrons from a heated filaments was passed two char… - Vidyadip

MCQ

A stream of electrons from a heated filaments was passed 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 :

A
$\sqrt {meV}$
✓
$\sqrt {2meV}$
C
$meV$
D
$2meV$

✓

Answer

Correct option: B.

$\sqrt {2meV}$

b
Kinetic energy $=e V=\frac{1}{2} m u^{2}$

$u=\sqrt{\frac{2 e V}{m}}$

Here, u is the velocity of the electron.

The de broglie wavelength is $\lambda=\frac{h}{m u}$

$\frac{h}{\lambda}=m u$

Substitute equation (1) in equation (2)

$\frac{h}{\lambda}=m \sqrt{\frac{2 e V}{m}}=\sqrt{2 m e V}$

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