Question

03 question (3 points) astronomers have detected hydrogen atoms in interstellar space in the n=740 excited state. suppose an atom in this excited state undergoes a transition from n=740 to n=731. 2nd attempt part 1 (1 point) what is the atom’s change in energy as a result of this transition? j see periodic table see hint

Explanation:

Step1: Recall the formula for energy levels in hydrogen atom

The energy of an electron in the \(n\)-th energy level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{2.178 \times 10^{-18}\ \text{J}}{n^2} \]
where \(n\) is the principal quantum number.

Step2: Calculate the energy at \(n = 740\)

Substitute \(n = 740\) into the formula:
\[ E_{740} = -\frac{2.178 \times 10^{-18}\ \text{J}}{(740)^2} \]
\[ E_{740} = -\frac{2.178 \times 10^{-18}\ \text{J}}{547600} \]
\[ E_{740} \approx -3.977 \times 10^{-24}\ \text{J} \]

Step3: Calculate the energy at \(n = 731\)

Substitute \(n = 731\) into the formula:
\[ E_{731} = -\frac{2.178 \times 10^{-18}\ \text{J}}{(731)^2} \]
\[ E_{731} = -\frac{2.178 \times 10^{-18}\ \text{J}}{534361} \]
\[ E_{731} \approx -4.076 \times 10^{-24}\ \text{J} \]

Step4: Calculate the change in energy \(\Delta E\)

The change in energy is given by \(\Delta E = E_{\text{final}} - E_{\text{initial}}\). Here, the initial state is \(n = 740\) and the final state is \(n = 731\). So:
\[ \Delta E = E_{731} - E_{740} \]
\[ \Delta E = (-4.076 \times 10^{-24}\ \text{J}) - (-3.977 \times 10^{-24}\ \text{J}) \]
\[ \Delta E = -9.9 \times 10^{-26}\ \text{J} \]
(Note: The negative sign indicates that the atom loses energy as it transitions to a lower energy level.)

Answer:

\(\boxed{-9.9 \times 10^{-26}\ \text{J}}\)