What is the de Broglie wavelength associated with an elector accelera…

The correct answer is Option (4) → 0.102 nm ##

The de Broglie wavelength ($\lambda$) for an electron accelerated through a potential $V$ is given by:

$\lambda = \frac{h}{\sqrt{2mqV}}$

By substituting the standard values for an electron ($h$ = Planck's constant, $m$ = mass of electron, $q$ = charge of electron), we get the simplified formula:

$\lambda = \frac{12.27}{\sqrt{V}}Å \quad \text{or} \quad \lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$

Calculation for $V = 144 V$:

First, find the square root of the voltage: $\sqrt{144} = 12$.

Substitute this into the formula:

$\lambda = \frac{1.227}{12} \text{ nm}$

$\lambda = 0.10225 \text{ nm}$

Rounding to three decimal places, we get 0.102 nm.