Question

Without doing detailed calculations, indicate which of the following electron transitions in the hydrogen atom results in the emission of light of the longest wavelength. Explain why.

a. from n = 1 to n = 2

b. from n = 4 to n = 3

c. from n = 1 to n = 6

d. from n = 3 to n = 2

Answer 1

The wavelength of the emitted radiation can be calculated using the Rydberg's Equation:

1/λ = R (1/n₁²   -   1/n₂²)                                 ;where λ= wavelength, R = Rydberg's constant, n₁ = the orbit                                                                          where    electron jumps, n₂ = the orbit from where electron                                                                          jumps

When λ is highest, (1/ λ) is lowest. When (1/ λ) is lowest, value of (1/n₁²   -   1/n₂²) is lowest. So we have to just calculate (1/n₁²   -   1/n₂²) part.

a. (1/n₁²   -   1/n₂²) = (1/2²   -   1/1²) = -0.75              (Negative sign indicates that energy will be absorbed                                                                                                          instead of emission of energy)

b. (1/n₁²   -   1/n₂²) = (1/3²   -   1/4²) = 0.049

c. (1/n₁²   -   1/n₂²) = (1/6²   -   1/1²) = - 0.972           (Negative sign indicates that energy will be absorbed                                                                                                          instead of emission of energy)

d. (1/n₁²   -   1/n₂²) = (1/2²   -   1/3²) = 0.139

We have to consider only the positive values. Value of (1/n₁²   -   1/n₂²)       is lowest for 'b'.

Answer: b