The ionisation potential of `H`-atom is `13.6 eV`. When it is excited from ground state by monochromatic radiations of `970.6 Å`, the number of emission lines will be (according to Bohr's theory)

Options

A. 10
B. 8
C. 6
D. 4

Answer

C

Explanation

`(1)/(lamda)= R [(1)/(n_(1)^(2))- (1)/(n_(2)^(2))]`
`rArr (1)/(970.6 xx 10^(-10)) = 1.097 xx 10^(7) [(1)/(1^(2))- (1)/(n_(2)^(2))]`
`rArr n_(2)= 4`
Number of spectral line `= (n(n-1))/(2)= (4 xx 3)/(2)= 6`