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

Given Information

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

Formula

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

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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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

Step by step solution

Given Information

Formula

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

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