The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of `108.5mm`. The ground state energy of an electron of this ion will e

Options

A. `3.4 eV
B. 13.6eV
C. 40.8eV
D. 27.2 eV

Answer

C

Explanation

for third line of balmer series `n_1 = 2, n_2 =5`
` thererfore (1) /( lamda ) = RZ^2 [ (1)/(n_(1)^(2)) - (1)/( n_(2)^(2))] ` gives `z^2 = ( n_(1)^(2) n_(2)^(2))/((n_(2)^(2) -n_(1)^(2) ) lamda R)`
on putting values `Z=2`
from ` E =- (13.6 Z^2 )/( n^2) = ( - 13.6 (2)^2)/((1)^2) =- 54.4 eV `