TG EAMCET 3 May 2025 Shift 2 Paper

In the atomic spectrum of hydrogen, the wavenumber corresponding to an electronic transition is given by the Rydberg formula.
v=RH(‌

1

n12

−‌

1

n22

)
For transition (i) : n=4⟶n=2 the corresponding wavelength is λ1.
For transition (ii) : n=3⟶n=1, the corresponding wavelength is λ2.
Using the Rydberg formula
‌‌

1

λ1

=RH(‌

1

22

−‌

1

42

)=RH(‌

1

4

−‌

1

16

)
‌=RH(‌

3

16

)
‌‌

1

λ2

=RH(‌

1

12

−‌

1

32

)=RH(1−‌

1

9

)=RH(‌

8

9

)
We now calculate the difference in wavelengths
λ1−λ2‌=‌

1

v1

−‌

1

v2

‌=‌

1

(‌ 3RH 16 )

−‌

1

(‌ 8RH 9 )

=‌

16

3RH

−‌

9

8RH

Taking LCM,
λ1−λ2‌=‌

(16.8−9.3)

24RH

‌=‌

(128−27)

24RH

=‌

101

24RH

Thus, λ1−λ2=‌

1

RH

(‌

101

24

)
Therefore, the correct option is (c).