JEE Main 24 Feb 2021 Shift 2 Solved Paper

Let nf,ni be the final and initial orbit.
As we know that,
‌

1

λ

=109×107[‌

1

nf2

−‌

1

ni2

]
Now, checking for each option, we get
(a) ‌

1

λ

∝[‌

1

32

−‌

1

42

]=[‌

1

9

−‌

1

16

]=0.05 ... (i)
(b) ‌

1

λ

∝[‌

1

1

−‌

1

4

]=0.75 ... (ii)
(c) ‌

1

λ

∝[‌

1

16

−‌

1

25

]=0.0225 ... (iii)
(d) ‌

1

λ

∝[‌

1

4

−‌

1

9

]=0.14 ... (iv)
The option (b) has highest value.
Since, frequency, f=‌

C

λ

⇒f∝‌

1

λ

∴ Frequency will be maximum for transition n=2 to n=1.