Question

If an electrons is accelerated through potential difference ${}^{\prime }{V}^{\prime }$ then De-Broglie wavelength associated with it is directly proportional to:

• A. $V$
• B. $\frac{1}{{}^3\sqrt{V}}$
• C. $\sqrt{V}$
• D. $\frac{1}{\sqrt{V}}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{V}}$

De-broglie wavelength of an electron is given by,

$\lambda =\frac{h}{p}$ ....(1)

Where p is the momentum of the electron. Momentum is related to the kinetic energy by the equation, $p=\sqrt{2Km}$

and a charge q when accelerated by a potential difference $V$ gains a kinetic energy $K=qV$.

Combining all these relations Eq.(1) , can be written as,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2Km}}=\frac{h}{\sqrt{2qVm}}$

Therefore, $\lambda$ $\alpha$ $\frac{1}{\sqrt{V}}$