Question

Which of the following transitions in hydrogen atoms emit photons of highest frequency?
\[\begin{align}
  & \text{A}\text{. n = 1 to n = 2} \\
 & \text{B}\text{. n = 2 to n = 6} \\
 & \text{C}\text{.n = 6 to n = 2} \\
 & \text{D}\text{. n = 2 to n =1} \\
\end{align}\]

Answer 1

Hint: From the Bohr -Rutherford model of an atom, we know that the nucleus of the atom is present in the centre of the atom and the electrons revolve around the nucleus. This is similar to the solar system where the sun remains in the centre and the other planets revolve around the sun.

Formula used: $\Delta E=E_{i}-E_{f}=R\left(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right)$ and $E=h\nu$

Complete step by step answer:
Hydrogen atom is a simple atom, to which the Bohr -Rutherford model of an atom can be applied. We know that the Bohr -Rutherford model of an atom, gives the relationship of the emission spectrum due to a hydrogen atom.
The spectral lines are due to the excitation of the electrons from on energy level of the hydrogen atom to the other, and it is given as $\Delta E=E_{i}-E_{f}=R\left(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right)$ where $E_{i}$ is the initial energy of the hydrogen, when the electron is in $n_{i}$ state and $E_{f}$ is the final energy of the hydrogen, when the electron is in $n_{f}$ state. And $R$ is the Rydberg constant
 Since the transition of the electrons occurs in the visible region, then we can say that $E=h\nu$, where $h$ is the Planck’s constant and $\nu$ is the frequency of the light.
Clearly if $\nu$ is greater, then the energy difference is also high.
Then if $\Delta E$ must be higher, then, clearly, the square of the product of the states must be a small value, as $\Delta E\propto \dfrac{n_{i}^{2}-n^{2}_{f}}{(n_{i}n_{f})^{2}}$.
Then clearly option A or D is the answer.
But we also know that emission occurs only if an excited molecule from a higher state comes to a lower state. Thus option D is the answer.

So, the correct answer is “Option D”.

Note: We know that the change in energy $\Delta E$ is generally given as the difference between the energy in the final state and the initial state. But here notice that $\Delta E={{E}_{i}}-{{E}_{f}}$, this implies the excited molecule from a higher state comes to a lower state.