Question

the first emission line of hydrogen atomic spectrum in Balmer series appears at A.5R/36 B.3R/4 C.7R/144 D.9R/400

Answer 1

For the first emission line in Balmer series, n1 = 2 and n2 = 3.
Using Rydberg's formula:
ν¯ =R (1n21−1n22) cm−1 = R (122−132)cm−1 = R (14−19)cm−1 = 5R36cm−1

Hence, option (1) is correct.
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If you can't understand than see attached picture

Answer 2

Answer: The correct option is A.

Explanation: Balmer series in the atomic spectra of Hydrogen has n = 2.

We are asked to find the first emission line, which means that the next energy level has n = 3

Using Rydberg's Equation:

$\bar{\nu}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right )cm^{-1}$

where,

$\bar{\nu}=\text{Wave number}$

R = Rydberg's constant

$n_1^2=\text{Lower energy level}=2$

$n_2^2=\text{Higher energy level}=3$

Putting values in above equation, we get

$\nu=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right )cm^{-1}$

$\nu=\frac{5R}{36}cm^{-1}$

The correct option is A.