Answer:
0.541 nm
Explanation:
The condition for maxima is,
[tex]dsin\theta=m\lambda[/tex]
Here, m=0,1,2,.....
And d is the slit separation, m is the order of maxima, [tex]\lambda[/tex] is the wavelength.
Given that, the 17.3 eV electron posses a wavelength of
[tex]\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m[/tex]
And the order of maxima is [tex]m=1[/tex].
And the angle at which first order maxima occur is, [tex]\theta=33^{\circ}[/tex].
Put these values in maxima condition while solving for d.
[tex]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[/tex]
Therefore, the slit separation is 0.541 nm.
