Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer 1

For He⁺ ion, the wave number (`barv`)associated with the Balmer transition, n = 4 to n = 2 is given by:

`bar "v" = 1/lambda = "RZ"^2(1/"n"_1^2 - 1/"n"_2^2)` 

Where

n₁ = 2

n₂ = 4

Z = atomic number of helium

`bar "v" = 1/lambda  = "R"(2)^2(1/4 - 1/16)`

`= 4"R" ((4-1)/16)`

`=bar "v" = 1/lambda = (3"R")/4`

`=> lambda = 4/(3"R")`

According to the question, the desired transition for hydrogen will have the same wavelength as that of He⁺.

`=> "R"(1)^2[1/"n"_1^2 -1/"n"_2^2]  = (3"R")/4`

`[1/"n"_1^2 - 1/"n"_2^2] = 3/4 .....(1)`

By hit and trail method, the equality given by equation (1) is true only when

n₁ = 1 and n₂ = 2.

∴ The transition for n₂ = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He⁺ spectrum.