Question

According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

• A. $n=5$ to $n=3$
• B. $n=6$ to $n=1$
• C. $n=5$ to $n=4$
• D. $n=6$ to $n=5$

Answer 1

Answer: (D)

$n=6$ to $n=5$

$E=-13.6\frac{z^2}{n^2}eV$.

Thus $E\propto \frac{1}{n^2}$.

For the transition $n_2\to n_1$ we have,
$\Delta E\propto (\frac{1}{n_1^2}-\frac{1}{n_2^2})$
If we observe this relationship closely, we can observe more higher value of n, more less the value of E. Thus least energetic photon is observed in transition of $n=6$ to $n=5$. Thus answer is option D.