Question

Which of the following electron transitions in a hydrogen atom emits light of the shortest wavelength? Circle the correct answer and explain your answer. (No calculation needed.) A. n = 4 to n = 1 B. n = 2 to n = 1 C. n = 7 to n = 4 D. n = 4 to n = 2

Answer 1

Answer:

Transition from n = 4 to n = 1  corresponds to shortest wavelength.

Explanation:

         Process                   $\mid \Delta n\mid =\mid \frac{1}{(n)_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid$

             A                                         0.9375

             B                                          0.75

             C                                          0.0421

             D                                          0.1875

According to Rydberg equation for electronic transition in H-like atoms:

           $\frac{1}{\lambda }=R_{H}\mid (\frac{1}{n_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid$

where, $\lambda$ is wavelength of light emitted or absorbed, $R_{H}$ is Rydberg constant.

So, higher the value of $\mid \Delta n\mid$, lower will the corresponding wavelength of light.

Hence process A will be associated with shortest wavelength.