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Chapter 37: Q. 3 - Excercises And Problems (page 1081)

The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series. Calculate the wavelengths of the first four members of the Lyman series

Short Answer

Expert verified

$\Rightarrow W a v e l e n g t h s o f t h e f i r s t f o u r m e m b e r s o f t h e L y m a n s e r i e s - F o r n = 2 λ = 121.57 n m F o r n = 3 λ = 102.57 n m F o r n = 4 λ = 97.26 n m F o r n = 5 λ = 94.98 n m$

Step by step solution

01

Given Information

Given : The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series.

02

Formula

$W e k n o w t h e f o r m u l a f o r H y d r o g e n s p e c t r u m f o r a n y s e r i e s \frac{1}{m^{2}} - \frac{1}{n^{2}} = \frac{91.18 (n m)}{λ} W h e r e m = 1 (G i v e n) n = 2, 3, 4, . . . S o, \frac{1}{1^{2}} - \frac{1}{n^{2}} = \frac{91.18 (n m)}{λ}, L e t b e t h e e q u a t i o n (1)$

03

Now we will calculate the wavelengths of the first four members of the Lyman series

\Rightarrow F o r n = 2 P u t n = 2 i n e q u a t i o n (1) \frac{1}{1^{2}} - \frac{1}{2^{2}} = \frac{91.18 (n m)}{λ} 1 - \frac{1}{4} = \frac{91.18 (n m)}{λ} λ = \frac{91.18 (n m)}{0.75} = 121.5733 \approx 121.57 n m \Rightarrow F o r n = 3 P u t n = 3 i n e q u a t i o n (1) \frac{1}{1^{2}} - \frac{1}{3^{2}} = \frac{91.18 (n m)}{λ} \frac{1}{1} - \frac{1}{9} = \frac{91.18 (n m)}{λ} λ = \frac{91.18 (n m)}{0.8888} = 102.57 n m \Rightarrow F o r n = 4 P u t n = 4 i n e q u a t i o n 1 \frac{1}{1^{2}} - \frac{1}{4^{2}} = \frac{91.18 (n m)}{λ} λ = \frac{91.18 (n m)}{0.9375} = 97.2586 \approx 97.26 n m \Rightarrow F o r n = 5 P u t n = 5 i n e q u a t i o n (1) \frac{1}{1^{2}} - \frac{1}{5^{2}} = \frac{91.18 (n m)}{λ} λ = \frac{91.18 (n m)}{0.96} = 94.9791 \approx 94.98 n m

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