an electron in a hydrogen atom moves from level 3 to level 1. in a second hydrogen atom, an electron drops…

Answer:

A. The first atom emits light with more energy.

Explanation:

Step1: Recall energy - level formula

The energy of an electron transition in a hydrogen - atom is given by $E = h\nu=E_i - E_f$, where $E_i$ is the initial energy level and $E_f$ is the final energy level. The energy levels of a hydrogen atom are given by $E_n=-\frac{13.6}{n^{2}}\text{ eV}$, where $n$ is the principal quantum number.

Step2: Calculate energy for first transition

For the electron moving from $n = 3$ to $n = 1$: $E_1=-\frac{13.6}{3^{2}}\text{ eV}=-\frac{13.6}{9}\text{ eV}\approx - 1.51\text{ eV}$ and $E_2 =-\frac{13.6}{1^{2}}\text{ eV}=-13.6\text{ eV}$. The energy of the emitted photon $\Delta E_1=E_1 - E_2=-\frac{13.6}{9}-(-13.6)=13.6\left(1 - \frac{1}{9}\right)=13.6\times\frac{8}{9}\approx12.09\text{ eV}$.

Step3: Calculate energy for second transition

For the electron moving from $n = 2$ to $n = 1$: $E_3=-\frac{13.6}{2^{2}}\text{ eV}=-\frac{13.6}{4}=-3.4\text{ eV}$ and $E_2=-13.6\text{ eV}$. The energy of the emitted photon $\Delta E_2=E_3 - E_2=-\frac{13.6}{4}-(-13.6)=13.6\left(1-\frac{1}{4}\right)=13.6\times\frac{3}{4} = 10.2\text{ eV}$.

Step4: Compare energies

Since $\Delta E_1\approx12.09\text{ eV}$ and $\Delta E_2 = 10.2\text{ eV}$, the first atom (electron transitioning from $n = 3$ to $n = 1$) emits light with more energy.