Respuesta :
Answer:
Transition from n = 4 to n = 1 corresponds to shortest wavelength.
Explanation:
Process [tex]\mid \Delta n\mid =\mid \frac{1}{(n)_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid[/tex]
A 0.9375
B 0.75
C 0.0421
D 0.1875
According to Rydberg equation for electronic transition in H-like atoms:
[tex]\frac{1}{\lambda }=R_{H}\mid (\frac{1}{n_{final}^{2}}-(\frac{1}{n_{initial}^{2}})\mid[/tex]
where, [tex]\lambda[/tex] is wavelength of light emitted or absorbed, [tex]R_{H}[/tex] is Rydberg constant.
So, higher the value of [tex]\mid \Delta n\mid[/tex], lower will the corresponding wavelength of light.
Hence process A will be associated with shortest wavelength.
