Question

The shortest wavelength present in the Brackett series of spectral line is –

• A. 912 Å
• B. 8201 Å
• C. 1.46 µ m
• D. 2.28 µ m

Answer 1

Answer: (C)

1.46 µ m

Answer 2

$\frac{1}{\lambda_{\min}}$= R(1)²$\left( \frac{1}{4^{2}} - \frac{1}{\infty^{2}} \right)$

Ž l_min = $\frac{16}{R}$= 16 × 912 Å