What will be the de Broglie wavelength of the electron accelerated by an electric potential of V volts? A λ=1.23/m B λ=1.23/h meters C λ=1.23/V nanometers D λ=1.23/V

The correct answer is:

C

The expression for the de Broglie wavelength of an electron accelerated by an electric potential of $\mathrm{V}$ volts is given by:

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

Given that $h$ is Planck's constant, $e$ is the charge of the electron, $V$ is the accelerating potential, and $m$ is the mass of the electron.

Using the given constants and simplifying, we get:

$$\lambda = \frac{1.23}{\sqrt{V}} , \text{nm}$$

Therefore, the correct option is:

C. $\lambda = \frac{1.23}{\sqrt{V}} , \text{nm}$