BITSAT 2024 Shift 1 Paper

To determine which transitions of a He+ion can produce a spectral line with the same wavelength as a given transition in a hydrogen atom, let's examine the Rydberg formula:
For a hydrogen atom ( Z=1 ), the wavelength λ1 for a transition from n2 to n1 is given by:
‌

1

λ1

=R(‌

1

n12

−‌

1

n22

)
Here, R is the Rydberg constant, and n1 and n2 are the principal quantum numbers with n2>n1.
For a He+ion (with Z=2 ), if the electron transitions from n4 to n3, the wavelength λ2 is given by:
‌

1

λ2

=R⋅4(‌

1

n32

−‌

1

n42

)
Assuming λ1=λ2, we have:
R(‌

1

n12

−‌

1

n22

)=R⋅4(‌

1

n32

−‌

1

n42

)
By comparing both sides, we get the relationships:
n1=‌

n3

2

,‌‌n2=‌

n4

2

Both n1,n2,n3, and n4 must be integers, as they represent principal quantum numbers.
This condition is satisfied when n3=2 and n4=4, ensuring that transitions corresponding to these quantum numbers yield the same wavelength for both He+and hydrogen.