Question

The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be(a) 3.4 eV(b) 13.6 eV(c) 54.4 eV(d) 122.4 eV

Answer 1

$<b>$

Option C is correct

Answer 2

Answer:

C) 54.4eV

Explanation:

Wavelength of hydrogen atom = 108.5nm (Given)

For third line of Balmer series, n1=2, n2=5

Therefore,

=  1/λ=RZ² [1/n²1-1/n²2]

= Z² = n²1n²2/ ( n²1n²2) λR

Putting the values Z = 2, hence the binding energy of electron in the ground state if ion is -  

E = - 13.6Z²/n²

E = - 13.6Z²/ 1²

E = - 54.4eV

Thus, the ground state energy of an electron of the ion will be 54.4eV