c
(c)

To get six possible emission lines the electron must be excite to the third level as

\(6=\frac{n}{2}(n+1)\)

\(\therefore n=3\)

Frequency of incident radiation is such that \(h f=13.6\left(1-\frac{1}{3^2}\right)\)

\(\text { or } \frac{h c}{\lambda}=13.6 \times \frac{8}{9} \quad \text { or } \quad \lambda=\frac{h c \times 9}{13.6 \times 8}\)

\(\text { or } \lambda=970 \,\mathring A\)