JEE Advanced 2021 Paper 1

From formula,
‌

1

λ

=R[‌

1

n12

−‌

1

n22

]
For Balmer series, n1=2
∴‌‌ For longest wavelength, transition occurs from n=3 to n=2.
∴‌

1

λmax

=R[‌

1

22

−‌

1

32

]
for shortest wavelength transition occurs from n=∞ to n=2
∴‌‌‌

1

λmin

=R[‌

1

22

−‌

1

∞2

]
∴‌‌‌

λ‌longest ‌

λ‌shorts ‌

=‌

9

5

For Paschen series, n1=3
λ‌longest ‌ of Balmer =‌

36

5R

and
λ‌shortest ‌ of Paschen =‌

9

R

Hence there is no overlap between the wavelength ranges of Balamer and Paschen series.
For Lyman series, n1=1
‌

1

λ

=R[‌

1

1

−‌

1

m2

]
Also ‌

1

λ0

=R
∴‌‌‌

1

λ

=‌

1

λ0

[1−‌

1

m2

]⇒λ=‌

λ0

1−‌ 1 m2

λ‌longest ‌ of Lyman =‌

4

3R

and λ‌shortest ‌ of Balmer =‌

4

R

Hence that wavelenght ranges of Lyman and Balmer series do not overlap.