IIT JEE Advanced 2011 Paper 1

For hydrogen-like atom, the wavelength emitted in Balmer series is given by

1

λ

= Z2R(

1

22

−

1

n2

)
The given wavelength of first line of hydrogen is 6561 Å, that is,

1

6561×10−10m

= R(

1

4

−

1

9

)
⇒ R = (

36

5×6561×10−10

)m−1
Therefore, for singly ionized helium, we get

1

λ

= 22R(

1

22

−

1

42

) = R(1−

1

4

)
=

36×3

4×5×6561×10−10m

⇒ λ = 1215 Å