Question

Shortest wavelength in Lyman series has wavelength of 915 Å. Longest wavelength of Balmer series has a value of?

(1) 5296 Å 

2) 3647 Å 

(3) 6588 Å 

(4) 7294 Å

Answer 1

Correct Option is (3) 6588 Å  

Lyman : \(\frac{1}{915} = RZ^2 (\frac{1}{1}- \frac{1}{\alpha})\)

              \(RZ^2 = \frac{1}{915}\) 

Balmer : Transition from n = 3 to n = 2

\(\frac{1}{\lambda}= RZ^2(\frac{1}{2^2}-\frac{1}{3^2})\) 

\(\frac{1}{\lambda}=\frac{1}{915}(\frac{5}{36})\) 

\(\lambda\) = 6588 Å