Amonoenergetic 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?

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=

Δy=λ(L/a)

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

so for λ

λ=Δy(a/L)

λ=4.2×10⁻¹¹m

The momentum of one of these electrons is then

p=h/λ

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

Convert joules into electronvolt we get

KE=854ev

b.a device that inputs heat into a cool object

conduction, convection, radiation