In the Balmer series of the hydrogen atom, the electron transition occurs from a higher energy level ( n≥3 ) to the second energy level ( n=2 ). The wavelength ( λ ) for these transitions is given by the Rydberg formula: ‌
1
λ
=R(‌
1
22
−‌
1
n2
) For different n values, this can be used to calculate the wavelength: For n=3 ‌
1
λ
=R(‌
1
4
−‌
1
9
)=R(‌
9−4
36
)=‌
5R
36
Therefore, λ=‌
36
5R
For n=4 ‌
1
λ
=R(‌
1
4
−‌
1
16
)=R(‌
16−4
64
)=‌
3R
16
Therefore, λ=‌
16
3R
For n=5 ‌
1
λ
=R(‌
1
4
−‌
1
25
)=R(‌
25−4
100
)=‌
21R
100
Therefore, λ=‌
100
21R
For n=6 ‌
1
λ
=R(‌
1
4
−‌
1
36
)=R(‌
36−4
144
)=‌
32R
144
=‌
8R
36
Therefore, λ=‌
36
8R
=‌
9
2R
Clearly, the correct option, based on our calculations with the transitions that correspond to the Balmer series, is from the option for n=5 :
