Answer
(a) $f=1.040727\times10^{13}Hz$
(b) Infrared region of electromagnetic spectrum.
Work Step by Step
From equation 29.6 momentum of photon is given by
$p=\frac{h}{\lambda}$
given that $p=2.3\times10^{-29}kg.m/s$, $h=6.63\times10^{-34}J.s$
so $\lambda=\frac{h}{p}$
so $\lambda=\frac{6.63\times10^{-34}J.s}{2.3\times10^{-29}kg.m/s}=2.882608\times10^{-5}m$
$\lambda=28826.08\times10^{-9}m=28826.08nm\approx28826nm$
(a) calculation of frequency
now since $f=\frac{c}{\lambda}$
putting $c=3\times10^{8}m/s$, $\lambda= 2.8826\times10^{-5}m$
$f=\frac{c}{\lambda} =\frac{3\times10^{8}m/s}{2.8826\times10^{-5}m}=1.040727\times10^{13}/s$
or $f=1.040727\times10^{13}Hz$
(b) Frequency suggests that it belongs to infrared region of the electromagnetic spectrum.
