The first member of the Balmer series of hydrogen atom has a wavelenght of `6561 Å` . The walvenght of the second member of the Balmer series (in nm) is ______.

Options

A. `13122 Å`
B. `3280 Å`
C. `4860 Å`
D. `2187 Å`

Answer

C

Explanation

As `(1)/(lamda)= R [(1)/(2^(2))- (1)/(n^(2))]` [For Balmer series]
`rArr (lamda_(2))/(lamda_(1))= ([(1)/(4)- (1)/(9)])/([(1)/(4)- (1)/(16)]) = (16)/(3) xx (5)/(36)= (20)/(27)`
`lamda_(2)= (20)/(27) xx 6561= 4860 Å`