Question

A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at an angle of 33.0°, what is the slit separation d (in nm)?

Answer 1

Answer:

0.541 nm

Explanation:

The condition for maxima is,

$dsin\theta=m\lambda$

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, $\lambda$ is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

$\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m$

And the order of maxima is $m=1$.

And the angle at which first order maxima occur is,  $\theta=33^{\circ}$.

Put these values in maxima condition while solving for d.

$d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm$

Therefore, the slit separation is 0.541 nm.