An electron of hydrogen atom on an excited state is having energy $$E_n = -0.85 \text{ eV}$$. The maximum number of allowed transitions to lower energy level is _______.

The energy of an electron in the $$n$$-th level of a hydrogen atom is given by $$E_n = -\frac{13.6}{n^2} \text{ eV}$$. When $$E_n = -0.85 \text{ eV}$$, we have $$-0.85 = -\frac{13.6}{n^2}$$ which leads to $$n^2 = \frac{13.6}{0.85} = 16$$ and hence $$n = 4$$, so the electron occupies the $$n = 4$$ state.

From $$n = 4$$, the electron can transition to any lower energy level $$n = 3, 2, 1$$. The maximum number of allowed transitions (spectral lines) is $$\binom{n}{2} = \binom{4}{2} = \frac{4!}{2! \times 2!} = 6$$, corresponding to the transitions $$4 \to 3$$, $$4 \to 2$$, $$4 \to 1$$, $$3 \to 2$$, $$3 \to 1$$, and $$2 \to 1$$. Therefore, the maximum number of allowed transitions is $$6$$.