Question

6. Minimum wavenumber possible for the spectral line present in Balmer series is [NCERT Pg. 45] (1) 82257 cm-1 (2) 109677 cm (3) 15233 cm (4) 18347 cm-1​

Answer 1

Answer:

(2) is the correct option

Answer 2

We have to find the minimum wavenumber for the spectral line present in Balmer series.

solution : for Blamer series, n₁ = 2 and n₂ = 3 , 4 , 5 , ...

from Rydberg's wave equation, $\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

to get minimum wavenumber, n₂ = 3

so, the wavenumber = $\frac{1}{\lambda}$ = RZ²(1/2² - 1/3²) = RZ²(1/4 - 1/9) = 5RZ²/36

R = Rydberg's constant = 1.0973 × 10⁷ m⁻¹

Z = atomic no [ for hydrogen , Z = 1]

now, wavenumber = 5(1.0973 × 10⁷ m)/36 = 1.5233 × 10⁶ m⁻¹

therefore the minimum wavenumbr possible for the spectral line present in Balmer series for hydrogen is  1.5233 × 10⁶ m⁻¹