Find the: (a) Maximum frequency, and

Solution:

Electron potential, V = 30 kV = 3 × 104 V

Hence, electron energy, E = 3 × 104 eV

Where, e = Charge on one electron = 1.6 × 10-19 C

(a) Maximum frequency by the X-rays = ν

The energy of the electrons:

E = hν

Where,

h = Planck’s constant = 6.626 × 10-34 Js

Therefore, $v=\frac{E}{h}$

$=\frac{1.6 \times 10^{-19} \times 3 \times 10^{4}}{6.626 \times 10^{-34}}=7.24 \times 10^{18} \mathrm{~Hz}$

Hence,  7.24 x 1018 Hz is the maximum frequency of the X-rays.

 

(b) The minimum wavelength produced:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{7.24 \times 10^{18}}=4.14 \times 10^{-11} \mathrm{~m}=0.0414 \mathrm{~nm}$