# 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

## Question

What will be the de Broglie wavelength of the electron accelerated by an electric potential of $\mathrm{V}$ volts?

A $\lambda=\frac{1.23}{\sqrt{m}}$

B $\quad \lambda=\frac{1.23}{\sqrt{h}} \, m$

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

D $\quad \lambda=\frac{1.23}{V}$

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## Answer

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

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