COMEDK Exam 2015 Solved Paper

Wavelength λ of spectral lines in H-atom is given as
‌

1

λ

=R(‌

1

n12

−‌

1

n22

)
For shortest wavelength of Paschen series,
n1=3‌ and ‌n2=∞
∴ From Eq. (i), we get
‌‌

1

λPS

=R(‌

1

32

−‌

1

∞

)=‌

R

9

⇒‌‌λPS‌=‌

9

R

. . . (ii)
For shortest wavelength in Balmer series,
n1=2,n2=∞
∴ From Eq. (i), we get
‌

1

λBS

‌=R(‌

1

22

−‌

1

∞

)=‌

R

4

⇒‌‌λBS‌=‌

4

R

. . . (iii)
For shortest wavelength of Lyman series,
n1=1‌ and ‌n2=∞
∴ From Eq. (i), we get
‌‌

1

λLS

=R(‌

1

12

−‌

1

∞

)=R
‌⇒‌‌λLS=‌

1

R

Hence, from Eq. (i), (ii) and (iii), we have
λPS:λBS:λLS‌=‌

9

R

:‌

4

R

:‌

1

R

‌=9:4:1