Answer
(a) The letter $f$ describes the atom.
(b) The atom's minimum possible energy is $-0.85~eV$
Work Step by Step
(a) We can find the value of $l$:
$L = \sqrt{l~(l+1)}~\hbar = 3.65\times 10^{-34}~J~s$
$\sqrt{l~(l+1)} = \frac{3.65\times 10^{-34}~J~s}{1.0546\times 10^{-34}~J~s}$
$\sqrt{l~(l+1)} = 3.46$
$(l)~(l+1) = (3.46)^2$
$(l)~(l+1) = 12$
$(l)~(l+1) = (3)(4)$
$l = 3$
When $l = 3$, then the letter $f$ describes the atom.
(b) When $l = 3$, then the minimum value of $n$ is $n = 4$
We can find the energy when $n = 4$:
$E = -\frac{13.60~eV}{n^2}$
$E = -\frac{13.60~eV}{4^2}$
$E = -0.85~eV$
The atom's minimum possible energy is $-0.85~eV$
