What transition in the He * spectrum would have the same wavelength as the first Lyman transition of hydrogen n 2 ton 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.

Text Solution

`2 rarr 1``5 rarr 3``4 rarr 2``6 rarr 4`

Answer : C

Solution : `bar(v)_(H) = R[(1)/(1^(2))-(1)/(2^(2))] cm^(-1)` ...(i) <br> `bar(v)_(He^(+)) = R(2)^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` ....(ii) <br> Now, `bar(v)_(He^(+)) = R[(1)/(1^(2))-(1)/(2^(2))] = R xx ((2)^(2))/((2)^(2)) [(1)/(1^(2))-(1)/(2^(2))]` <br> `= R(2)^(2)[(1)/(2^(2))-(1)/(4^(2))]` ...(iii) <br> Comparing (ii) and (iii), it is clear that <br> `n_(2) = 4, n_(1) = 2`.