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Grade: 12 

                        
which one of the following transitions in a hydrogen atom will emit the photon of shortest wavelength n = 2 to n = 1 n = 6 to n = 5 n = 6 to n = 1 n = 5 to n = 4
2 years ago

Answers : (2)

Avni Chauhan
askIITians Faculty
247 Points 
							shortest wavelength implies maximum energy difference. Maximum energy difference comes when electrons are transferred from n = 2 to n = 1

energy in any orbital = -13.6/ n2
2 years ago
Susmita
425 Points 
							
wave number
\nu =\frac{1}{\lambda} = RZ^2 (\frac{1}{n`^2}-\frac{1}{n^2}) 
Shortest wavelength will correspond to largest \nu.
For n=2 to n=1,
\nu = RZ^2 (\frac{1}{1^2}-\frac{1}{2^2})=0.75RZ^2
Similarly for n=6 to n=5,
 
\nu = RZ^2 (\frac{1}{5^2}-\frac{1}{6^2})=0.012RZ^2
For n=6 to n=1
\nu = RZ^2 (\frac{1}{1^2}-\frac{1}{6^2})=0.972RZ^2
For n=5 to 4
\nu = RZ^2 (\frac{1}{4^2}-\frac{1}{5^2})=0.225RZ^2
So n=6 to 1 is the answer.
Please approve the answer if you are helped.
2 years ago
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