Question

Which of the following transitions in a hydrogen atom, will emit photon of highest frequency?

• A. $n=1\to n=2$
• B. $n=2\to n=1$
• C. $n=2\to n=6$
• D. $n=6\to n=2$

Answer 1

The correct option is B $n=2\to n=1$

Out of the four options, we will check only for option $\left(B\right)$ and $\left(D\right)$, because photon is emitted only when electron jumps from higher orbit to lower orbit.


Energy of an electron in $n^{\text{th}}$ orbit is,
$E_n=\frac{E_1}{n^2}(E_1\to \text{energy of the electron in first orbit})$

Let, $E_1=E\Rightarrow E_2=\frac{E}{4}\text{and}{E}_6=\frac{E}{36}$

Energy of the photon emitted in the $n=2\to n=1$ transition is,
$\Delta E=|E_2-E_1|=|\frac{E}{4}-E|=\frac{3E}{4}$

And, Energy of the photon emitted in the $n=6\to n=2$ transition is,
$\Delta E=|E_6-E_2|=|\frac{E}{36}-\frac{E}{4}|=\frac{2E}{9}$

As energy difference between $n=1$ and $n=2$ is more, as compared to the energy difference between $n=6$ to $n=2$ , and, we know that,
$\Delta E=h\nu \to \text{energy of the photon emitted}$

Hence, $\left(B\right)$ is the correct answer.