Question:
An atom has two energy levels with energy difference 2.2 eV between them. A gas of these atoms has \(8 \times 10^{20}\) atoms in the upper state and \(5 \times 10^{20}\) atoms in the lower state. Ignoring spontaneous emission, the maximum possible energy released by this gas of atoms by stimulated emission is \(E\) Joules. The value of \(E\) (rounded off to one decimal place) is ______ \((e = 1.6\times10^{-19}\text{ C})\)
An atom has two energy levels with energy difference 2.2 eV between them. A gas of these atoms has \(8 \times 10^{20}\) atoms in the upper state and \(5 \times 10^{20}\) atoms in the lower state. Ignoring spontaneous emission, the maximum possible energy released by this gas of atoms by stimulated emission is \(E\) Joules. The value of \(E\) (rounded off to one decimal place) is ______ \((e = 1.6\times10^{-19}\text{ C})\)
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Hint:
Stimulated absorption and stimulated emission share the same rate constant, so the net process stops once the populations equalize. Find how many atoms must move from the upper to the lower state for that to happen, then multiply by the photon energy.
Stimulated absorption and stimulated emission share the same rate constant, so the net process stops once the populations equalize. Find how many atoms must move from the upper to the lower state for that to happen, then multiply by the photon energy.
Updated On: Jul 28, 2026
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Correct Answer: 52.8
Solution and Explanation
Step 1: Understand what limits stimulated emission.
For a non degenerate two level system, the Einstein coefficients for stimulated absorption and stimulated emission are equal (\(B_{12}=B_{21}\)). So in a radiation field, atoms in the lower state absorb photons at a rate proportional to \(N_1\), while atoms in the upper state emit photons (stimulated) at the same rate constant times \(N_2\). The net number of photons produced per unit time is proportional to \(N_2 - N_1\).
Step 2: See why the process has a maximum.
Every net stimulated emission event moves one atom from the upper state to the lower state: \(N_2\) drops by 1 and \(N_1\) rises by 1, so the difference \(N_2-N_1\) drops by 2 each time. This keeps happening as long as \(N_2 > N_1\). Once \(N_2 = N_1\), emission and absorption occur at equal rates and cancel exactly, so no further net energy can be released, ignoring spontaneous emission. This equalization point is what caps the total energy release.
Step 3: Find how many atoms make the downward transition.
The initial difference is
\[ N_2 - N_1 = 8\times10^{20} - 5\times10^{20} = 3\times10^{20} \]
Since each net transition reduces this difference by 2, the number of atoms that can move down before the difference reaches zero is
\[ \Delta N = \frac{N_2-N_1}{2} = 1.5\times10^{20} \]
Step 4: Convert to energy.
Each downward transition releases one photon of energy 2.2 eV, so the total energy released is
\[ \Delta N \times 2.2\text{ eV} = 1.5\times10^{20}\times2.2 = 3.3\times10^{20}\text{ eV} \]
Converting to joules using \(e=1.6\times10^{-19}\) C,
\[ E = 3.3\times10^{20}\times1.6\times10^{-19} = 52.8\text{ J} \]
Final Answer:
Rounded to one decimal place, \[ \boxed{E = 52.8\text{ J}} \]
For a non degenerate two level system, the Einstein coefficients for stimulated absorption and stimulated emission are equal (\(B_{12}=B_{21}\)). So in a radiation field, atoms in the lower state absorb photons at a rate proportional to \(N_1\), while atoms in the upper state emit photons (stimulated) at the same rate constant times \(N_2\). The net number of photons produced per unit time is proportional to \(N_2 - N_1\).
Step 2: See why the process has a maximum.
Every net stimulated emission event moves one atom from the upper state to the lower state: \(N_2\) drops by 1 and \(N_1\) rises by 1, so the difference \(N_2-N_1\) drops by 2 each time. This keeps happening as long as \(N_2 > N_1\). Once \(N_2 = N_1\), emission and absorption occur at equal rates and cancel exactly, so no further net energy can be released, ignoring spontaneous emission. This equalization point is what caps the total energy release.
Step 3: Find how many atoms make the downward transition.
The initial difference is
\[ N_2 - N_1 = 8\times10^{20} - 5\times10^{20} = 3\times10^{20} \]
Since each net transition reduces this difference by 2, the number of atoms that can move down before the difference reaches zero is
\[ \Delta N = \frac{N_2-N_1}{2} = 1.5\times10^{20} \]
Step 4: Convert to energy.
Each downward transition releases one photon of energy 2.2 eV, so the total energy released is
\[ \Delta N \times 2.2\text{ eV} = 1.5\times10^{20}\times2.2 = 3.3\times10^{20}\text{ eV} \]
Converting to joules using \(e=1.6\times10^{-19}\) C,
\[ E = 3.3\times10^{20}\times1.6\times10^{-19} = 52.8\text{ J} \]
Final Answer:
Rounded to one decimal place, \[ \boxed{E = 52.8\text{ J}} \]
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