Solution :
Here Energy $E= [4.4 Kev = 14.4 \times 10^3 \times 1.6 \times 10^{-19}]$
$E= h \alpha$
$\alpha= \large\frac{E}{h}$
$\qquad= \large\frac{14.4 \times 10^3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\qquad= 3 \times 10^{11} \;MHZ$
Hence this frequency lies in the x-ray region of the e.m spectrum.