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))]`
`implies (1)/( 970.6 xx 10 ^(-10))= 1.097 xx 10^7 [(1)/(I^2 - (1)/(n_(2)^(2))) ]implies n_2 =4`
` therefore ` Number of emmission line `N= ( n(n-1))/( 2) = ( 4 xx 3 )/(2) = 6`