Question

The Balmer-Rydberg equation can be extended to ions with only one electron, such as He+. In that case it has the form: 1/λ = Z2R(1/m2 - 1/n2), where Z is the atomic number.

What is the energy of the photon required to promote an electron in He+ from a 1s orbital to a 2p orbital? A) (3/4)hcR
B) 3hcR
C) 4hcR
D) 12hcR

And please explain if possible

Answer 1

Ans :- Option (B) i.e. 3hcR is the correct answer.

Explanation :-

Atomic number of He = Z = 2

m = 1 and n = 2 for 1s and 2s orbitals respectively.

Substitute the values of Z, m and n in the given Balmer-Rydberg equation i.e.

1/λ = Z²R(1/m² - 1/n²)  

we have

1/λ = (2)² R [ 1/(1)² - 1/(2)² ]

1/λ = 4R [ 1 - 1/4 ]

1/λ = 4 R [ 3 / 4 ]

1/λ = 3 R

We know  

Energy of the photon (E) is :

E = hc / λ

Substitute the value of 1/λ in the above equation, we have

E = 3 hcR

Hence option (B) is the correct answer.