Question

1. answer the following questions regarding a h atom in which an electron moves from the fourth energy level back to the second energy level. for both questions, express your answers to three significant figures. a. calculate the energy change (in j) for the h atom. (7 pt)

Explanation:

Step1: Recall the energy - level formula for hydrogen atom

The energy of an electron in the $n$th level of a hydrogen atom is given by $E_n=-\frac{13.6}{n^{2}}\text{ eV}$. First, convert the energy from electron - volts to joules using the conversion factor $1\text{ eV}=1.6\times 10^{- 19}\text{ J}$.

Step2: Calculate energy at $n = 4$

$E_4=-\frac{13.6}{4^{2}}\text{ eV}=-\frac{13.6}{16}\text{ eV}=- 0.85\text{ eV}$. Converting to joules, $E_4=-0.85\times1.6\times 10^{-19}\text{ J}=-1.36\times 10^{-19}\text{ J}$.

Step3: Calculate energy at $n = 2$

$E_2=-\frac{13.6}{2^{2}}\text{ eV}=-\frac{13.6}{4}\text{ eV}=-3.4\text{ eV}$. Converting to joules, $E_2=-3.4\times1.6\times 10^{-19}\text{ J}=-5.44\times 10^{-19}\text{ J}$.

Step4: Calculate the energy change

The energy change $\Delta E=E_2 - E_4$. Substituting the values, $\Delta E=(-5.44\times 10^{-19}\text{ J})-(-1.36\times 10^{-19}\text{ J})=-5.44\times 10^{-19}\text{ J}+1.36\times 10^{-19}\text{ J}=-4.08\times 10^{-19}\text{ J}$. The negative sign indicates that energy is emitted. The magnitude of the energy change is $4.08\times 10^{-19}\text{ J}$.

Answer:

$4.08\times 10^{-19}\text{ J}$