Question

among the following h atom transitions, which would emit a photon of light with the greatest energy? answer: a n = 5 to n = 3 b n = 4 to n = 2 c n = 2 to n = 1 d n = 5 to n = 4 e n = 6 to n = 3

Explanation:

Step1: Recall energy - level formula

The energy of a photon emitted during an electron transition in a hydrogen - like atom is given by $\Delta E = E_i - E_f$, and $E_n=-\frac{13.6}{n^{2}}\text{ eV}$. The greater the difference in energy levels ($\Delta n$), the greater the energy of the emitted photon.

Step2: Calculate $\Delta n$ for each option

• Option A: $\Delta n=5 - 3=2$
• Option B: $\Delta n=4 - 2=2$
• Option C: $\Delta n=2 - 1=1$
• Option D: $\Delta n=5 - 4=1$
• Option E: $\Delta n=6 - 3=3$

Step3: Compare $\Delta n$ values

The larger the $\Delta n$, the larger the energy of the emitted photon. Since $3>2 > 1$, the transition with the greatest $\Delta n$ is from $n = 6$ to $n = 3$.

Answer:

E. $n = 6$ to $n = 3$