JEE Main 26 Feb 2021 Shift 1 Solved Paper

By using Rydberg's formula, ‌

1

λ

=R[‌

1

n22

−‌

1

n12

]
where, R is Rydberg constant.
For wavelength of third member of Lyman series n2=1 and n1=4.
∴‌‌‌

1

λ1

=R[‌

1

12

−‌

1

42

]=R[‌

1

1

−‌

1

16

] . . . (i)
For wavelength of first member of Paschen series, n2=3 and n1=4
∴‌‌‌

1

λ2

=R[‌

1

32

−‌

1

42

]=R[‌

1

9

−‌

1

16

] . . . (ii)
On dividing Eq. (ii) by Eq. (i), we get
‌

λ1

λ2

=‌

[ 1 9 − 1 16 ]

[‌ 1 1 −‌ 1 16 ]

=‌

7

9×15

=‌

7

135