Question

The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen will be

• A. 4215 Å
• B. 1437Å
• C. 3942Å
• D. 3647Å

Answer 1

Answer: (D)

3647Å

Answer 2

$\frac{1}{\lambda_{\text{shortest}}} = RZ^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$ $= 109678 \times 1^{2} \times \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right)$

$\lambda = 3.647 \times 10^{- 5}cm$ $= 3647Å$