Question

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 :

• A. $2meV$
• B. $\sqrt{meV}$
• C. $\sqrt{2meV}$
• D. meV

Answer 1

The correct option is C $\sqrt{2meV}$

Kinetic energy $=eV=\frac{1}{2}m{u}^2$

$u=\sqrt{\frac{2eV}{m}}$......(1)

Here, u is the velocity of the electron.

The de broglie wavelength is $\lambda =\frac{h}{mu}$.

$\frac{h}{\lambda }=mu$......(2)

Substitute equation (1) in equation (2)

$\frac{h}{\lambda }=m\sqrt{\frac{2eV}{m}}=\sqrt{2meV}$

Hence, the correct option is $C$