In a hydrogen atom, the electron is in nth excited it comes down to first excited state by emitting ten different wavlength . The value of n is

Answer

6

Explanation

When electron jumps from `n^(th)` state to ground state, number of possible emission lines `= n (n - 1)//2`, But here transition takes place from `n^(th)` state to state `n_(1) =2`
Here, number of possible emission lines `= n (n - 1) (n - 2)//2 = 10` (gives).