Answer
a) $p=1.36\times 10^{-27}Kgm/s$
b) $v=0.814m/s=81.4cm/s$
Work Step by Step
(a) We know that
$E_n=\frac{-13.6}{n^2}eV$
$\implies E_4=\frac{-13.6}{4^2}=-0.85eV$
and $E_2=\frac{-13.6}{2^2}=-3.4eV$
Now $E=-0.85-(-3.4)$
$E=2.55eV=4.08\times 10^{-19}J$
We can find the linear momentum as
$p=\frac{E}{c}$
We plug in the known values to obtain:
$p=\frac{4.08\times 10^{-19}}{3\times 10^8m/s}$
$p=1.36\times 10^{-27}Kgm/s$
(b) We know that
$v=\frac{p}{m}$
We plug in the known values to obtain:
$v=\frac{1.36\times 10^{-27}}{1.67\times 10^{-27}}$
$v=0.814m/s=81.4cm/s$
