Question

A hydrogen atom emits ultraviolet radiation of 102.5 nm. Then the quantum numbers of the states involved in the transition are –

• A. n =, n = 1
• B. n = 5, n = 1
• C. n = 3, n = 1
• D. None of these

Answer 1

Answer: (C)

n = 3, n = 1

Answer 2

n₁ = 1 (for U.V. radiation)

l = 1025 Å

$\frac{1}{\lambda}$= R(1)²$\left( \frac{1}{1^{2}} - \frac{1}{n_{2}^{2}} \right)$ Ž 1 –$\frac{1}{n_{2}^{2}}$=$\frac{1}{\lambda_{R}}$

$\sqrt{1 - \frac{1}{\lambda_{R}}}$=$\frac{1}{n_{2}}$ Ž n₂ =$\frac{1}{\sqrt{1 - \frac{1}{\lambda_{R}}}}$=$\frac{1}{\sqrt{1 - \frac{912}{1025}}}$ $\approx$ 3