Answer
(a) $-0.85eV$
(b) $3.657\times 10^{-34}Js$
(c) greater than
(d) greater than
Work Step by Step
(a) We know that
$Energy \space of\space the \space atom=\frac{-13.6}{n^2}eV$
$\implies Energy \space of\space the \space atom=\frac{-13.6}{(4)^2}=-0.85eV$
(b) We know that
$L=\sqrt{l(l+1)}\frac{h}{2\pi}$
$\implies L=\sqrt{3(3+1)}\frac{h}{2\pi}$
$\implies L=\frac{(1.732\times 2)(6.63\times 10^{-34})}{2\times 3.14}=3.657\times 10^{-34}Js$
(c) We know that the energy of atom 2 is greater than the energy of atom 1
As atom 1 is in 4f state, we have $n=4$
$\implies E_4=\frac{-13.6}{(4)^2}=-0.85eV$
Atom 2 is in 5d state, we have $n=5$
$\implies E_5=\frac{-13.6}{(5)^2}=-0.544eV$
Thus, $E_5\gt E_4$, which shows that atom 2 has greater energy as compared to atom 1.
(d) The angular momentum for the first atom is $L_1=\sqrt{3(3+1)}\frac{h}{2\pi}$
$\implies L_1=3.657\times 10^{-34}Js$
and the angular momentum for for the second atom is given as
$L_2=\sqrt{2(2+1)}\frac{h}{2\pi}=(2.236)(\frac{6.63\times 10^{-34}}{2\times 3.14})=2.36\times 10^{-34}Js$
Thus, we conclude that the angular momentum of the $f$ sub level is greater than the angular momentum of the $d$ sub level and atom 1 has greater angular momentum as compared to that of atom 2.
