Question

A tungsten target is struck by electrons that have been accelerated from rest through a 27.3-kV potential difference. Find the shortest wavelength of the radiation emitted.

Answer 1

4.555 x 10^-11 m

Step-by-step explanation:

Potential difference, V = 27.3 kV

Let the wavelength is λ.

The energy associated with the potential difference, E = 27.3 keV

E = 27.3 x 1000 x 1.6 x 10^-19 J = 4.368 x 10^-15 J

The energy associated with the wavelength is given by

`E=(hc)/(\lambda )`

Where, h is Plank's constant = 6.63 x 10^-34 Js

c is velocity of light 3 x 10^8 m/ s

By substituting the values, we get

`4.368*10^(-15)=(6.634*10^(-34)* 3* 10^(8))/(\lambda )`

λ = 4.555 x 10^-11 m