Question

Four lowest energy levels of H-atom are shown in the figure. The number of possible emission lines would be.

$\begin{array} { l l } & n = 4 \\ & n = 3 \\ & n = 2 \\ & n = 1 \end{array}$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (D)

6

Answer 2

Number of possible emission lines $= \frac{n(n - 1)}{2}$

Where n = 4; Number $= \frac{4(4 - 1)}{2} = 6.$