VITEEE 2013 Paper

We have to compare wavelength of transition in the H-spectrum with the Balmer transition n = 4 to n = 2 of He+ spectrum.
Since λH = λHe+
∴ RHZH2 [

1

n12

−

1

n22

] = RHZHe+2 [

1

22

−

1

42

]
1 × [

1

n12

−

1

n22

] = 4 × (

1

4

×

1

16

)
[

1

n12

−

1

n22

] = 4 ×

4−1

16

[

1

n12

−

1

n22

] =

3

4

If n1 = 1 , then n2 = 2 , 3 , ...
For first line n2 = 2 , n1 = 1
[

1

n12

−

1

n22

] =

1

1

−

1

4

=

3

4

Hence, trnasition n2 = 2 to n1 = 1 will give spectrum of the same wavelength as that of Balmer transition, n2 = 4 to n1 = 2 in He+.