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

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jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.