$E_n = – 0.278$ eV

$textbf{Find:}$ n

$textbf{Formula:}$

$$
begin{gather}
E_n = -(13.6 text{ eV}) times dfrac{1}{n^2}
end{gather}
$$

Using Eq (1), we can derive a formula to compute for n, such that:

$$
begin{align*}
n^2 &= dfrac{13.6 text{ eV}}{0.278 text{ eV}} \
n &= sqrt{dfrac{13.6 text{ eV}}{0.278 text{ eV}}} \
n &= 7
end{align*}
$$

Therefore, the Bohr orbit that has the energy of -0.278 eV is $boxed{text{n =7}}$

$E_1 = -0.72$ eV

$E_2 = -0.46$ eV

$textbf{Find:}$ n, such that:

$$
begin{gather}
n_1 < n < n_2
end{gather}
$$

$textbf{Formula:}$

$$
begin{gather}
E_n = -(13.6 text{ eV}) times dfrac{1}{n^2}
end{gather}
$$

Using (2), we can derive an equation that can calculate $n_1$ and $n_2$, such that:

$$
begin{align*}
n_1 &= sqrt{dfrac{-(13.6 text{ eV})}{-0.72 text{ eV}}} \
&= 4.36 \
n_2 &= sqrt{dfrac{-(13.6 text{ eV})}{-0.46 text{ eV}}} \
&= 5.43 \
end{align*}
$$

Using the relationship stated in (1):

$$
begin{gather*}
4.36 < n < 5.43
end{gather*}
$$

Therefore, the Bohr orbit whose energy lies in the range of $E_1$ and $E_2$ is $boxed{text{n = 5}}$