Find the

Potential of the electrons, $V=30 \mathrm{kV}=3 \times 10^{4} \mathrm{~V}$

Hence, energy of the electrons, $E=3 \times 10^{4} \mathrm{eV}$

Where,

e = Charge on an electron = 1.6 × 10−19 C

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

The energy of the electrons is given by the relation:

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, the maximum frequency of X-rays produced is $7.24 \times 10^{18} \mathrm{~Hz}$.

(b)The minimum wavelength produced by the X-rays is given as:

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

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

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.