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

0.541 nm

Step-by-step 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* 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=(1* 0.295* 10^(-9)m)/(sin33^(\circ)) \\d=(0.295* 10^(-9)m)/(0.545) \\d=0.541* 10^(-9)m}\\d=0.541 nm`

Therefore, the slit separation is 0.541 nm.