Verified Solution • 2025–26 Session moderate

QPages 384-386: Q2: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q2 Solution: Find the wavelength of the radiation emitted by hydrogen in the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently reverse the principal quantum numbers in the Rydberg formula, calculating $1/n_{upper}^2 - 1/n_{lower}^2$ instead of the correct $1/n_{lower}^2 - 1/n_{upper}^2$, which results in a negative wavelength. Always ensure the smaller $n$ value is in the first term to yield a positive wavenumber for emission.

Find the wavelength of the radiation emitted by hydrogen in the transitions a)

n = 3

to

n= 2

, b)

n = 5

to

n = 4

and c)

n = 10

to

n = 9

.

Step-by-Step Verified Solution

**Answer:** (a) 656 nm (b) 4050 nm (c) 38862 nm. **Step-by-Step Explanation:** Using Rydberg's formula:

\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

with

R = 1.097 \times 10^7

m__. (a) n_ = 2, n_ = 3:

\frac{1}{\lambda} = R(1/4 - 1/9) = 5R/36

.

\lambda = 36/(5 \times 1.097 \times 10^7) = 6.56 \times 10^{-7}

m = 656 nm. (b) n_ = 4, n_ = 5:

\frac{1}{\lambda} = R(1/16 - 1/25) = 9R/400

.

\lambda = 400/(9 \times 1.097 \times 10^7) = 4.05 \times 10^{-6}

m = 4050 nm. (c) n_ = 9, n_ = 10:

\frac{1}{\lambda} = R(1/81 - 1/100) = 19R/8100

.

\lambda = 8100/(19 \times 1.097 \times 10^7) = 3.886 \times 10^{-5}

m = 38862 nm [citation:3][citation:7].

All Questions in this Chapter

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