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If the shortest wave length of Lyman series of H atom is x, then the wave length of the first line Balmer series of H atom will be 1) 9x/52) 36x/53) 5x/94) 5x/36Explanation please..Thank You

If the shortest wave length of Lyman series of H atom is x, then the wave length of the first line Balmer series of H atom will be 1) 9x/52) 36x/53) 5x/94) 5x/36Explanation please..Thank You

Grade:11

2 Answers

Adityan
17 Points
7 years ago
 
The wave no. of any series is given by;
(1/\lambda ) = 109678 * [ (1/n^{2}_{1}) - (1/n_{2}^{2}) ]
where 109678 is a constant,lets say = \alpha;
and,the series starts from n1 to n2;
 
Here,x is the wavelength for the last line(shortest wavelength) in lyman series,
therefore,n​1=1 to n​2=\bowtie(infinity)​;
\Rightarrow    ​(1/x)=\alpha​​[(1/1) – ​(1/\bowtie)]    =   \alpha ​(​1 - 0) = \alpha
\Rightarrow    ​ \alpha =​ 1/x              
 
For wavelength of first line of balmer series,
n​1=2 to n​2=3  ​;lets say the wavelength to be found be = \lambda;​
\Rightarrow (1/\lambda ) = \alpha *[(1/2^{2}) - (1/3^{2})] = \alpha(5/36)
Substituting the value
\Rightarrow (1/\lambda ) = (5/36)(1/x) = 5/(36x) \Rightarrow \lambda =36x/5
Ans(2)
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ankit singh
askIITians Faculty 614 Points
3 years ago
 
For wavelength of first line of balmer series,
n​1=2 to n​2=3  ​;lets say the wavelength to be found be = \lambda;​
\Rightarrow (1/\lambda ) = \alpha *[(1/2^{2}) - (1/3^{2})] = \alpha(5/36)
 

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