Answer
(a) $1.09\times 10^{15}Hz; 275nm$
(b) Ultraviolet
Work Step by Step
(a) We know that
$f=\frac{E}{h}$
We plug in the known values to obtain:
$f=\frac{7.24\times 10^{-19}J}{6.623\times 10^{-34}J.s}$
$f=1.09\times 10^{15}Hz$
Now $\lambda=\frac{c}{f}$
We plug in the known values to obtain:
$\lambda=\frac{3\times 10^8m/s}{1.09\times 10^{15}Hz}$
$\lambda=275\times 10^{-9}m$
$\lambda=275nm$
(b) We know that the wavelength of the emitted photon is $275nm$. The range of the wavelength of the $UV$ region is $\lambda_{UV}\lt 400nm$. Thus, the photon lies in the ultra violet region of the electromagnetic spectrum.
