Question

The diagram shows the energy levels for an electron in a hydrogen atom. Which transition shown represents the emission of a photon with the most energy?

• A. $\text{III}$
• B. $\text{IV}$
• C. $\text{I}$
• D. $\text{II}$

Answer 1

The correct option is A $\text{III}$

Emission of a photon takes place when an electron makes a transition from higher energy state to lower energy state.

In transition $\text{I}$, the electron makes a transition from lower energy state to higher energy state, hence, there will not be emission of photon in transition $\text{I}$.

Energy of an orbital state in hydrogen atom is given as

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

So, for $n=1\to E_1=-13.6\text{eV}$

for $n=2\to E_2=-3.4\text{eV}$

for $n=3\to E_3=-1.51\text{eV}$

for $n=4\to E_4=-0.85\text{eV}$

In transition $\text{II}$,

$\Delta E_{II}=E_4-E_3=0.66\text{eV}$

In transition $\text{III}$,

$\Delta E_{III}=E_2-E_1=10.2\text{eV}$

In transition $\text{IV}$,

$\Delta E_{IV}=E_4-E_2=2.55\text{eV}$

$\Delta E$ is maximum for transition $\text{III}$

$\therefore$ Emission of a photon with maximum energy will be in transition $\text{III}$.

Hence, option $\left(\text{A}\right)$ is correct.