Question

In the Bohr hydrogen atom, the electronic transition emitting light of longest wave length is:

• A. n = 2 to n =3
• B. n = 4 to n = 3
• C. n = 3 to n = 2
• D. n = 2 to n = 1

Answer 1

Answer: (B)

In atoms, as the number of shell increases, the energy differences between the shells decreases. It means that the energy difference between $n_1$ and $n_2$ is higher than between $n_2$ and $n_3$ and in the same way energy difference is more between $n_2$ and $n_3$ than between $n_3$ and $n_4.$ We know that, $E=hv=\frac{hc}{\lambda }$ wave length is inversely proportional to energy difference. So, in the given alternates wave length is maximum between $n_4$ to $n_3$ because between $n_4$ and $n_3$ energy difference is minimum.