Question

A monoenergetic beam of electrons is incident on a single slit of width 0.500 nm. A diffraction pattern is formed on a screen 25.0 cm from the slit.

If the distance between successive minima of the diffraction pattern is 2.10 cm, what is the energy of the incident electrons?

Answer 1

Answer:

KE=1.366×10⁻¹⁶J or

KE=854ev

Explanation:

single slit diffraction pattern occur where sinФ=mλ/a for m=±1,±2,±3,.......

Here λ is wavelength

The spacing between successive minima is then

Δy=$y_{m+1}-y_{m}$

Δy=λ(L/a)

as given Δy=2.10cm, L=25cm and a=0.500nm

so for λ

λ=Δy(a/L)

 $=(2.10*10^{-2}m )(\frac{(0.500*10^{-9}m)}{25*10^{-2}m} )$

λ=4.2×10⁻¹¹m

The momentum of one of these electrons is then

p=h/λ

$p=\frac{6.63*10^{-34} }{4.2*10^{-11}}\\ p=1.578*10^{-23}kgm/s$

Assuming the electron is non-relativistic, its kinetic energy is

$KE=p^{2}/2m_{e}\\  KE=\frac{(1.578*10^{-23} )^{2} }{2(9.11*10^{-31})}\\ KE=1.366*10^{-16}J$

Convert joules into electronvolt we get

KE=854ev