Question:
A photon of energy 12.09 eV is absorbed by a hydrogen atom in its first excited state, resulting in the ejection of an electron. The kinetic energy (in J) of the emitted photoelectron is ($1 eV = 1.6 \times 10^{-19}$ J, $R_H = 13.6$ eV):
A photon of energy 12.09 eV is absorbed by a hydrogen atom in its first excited state, resulting in the ejection of an electron. The kinetic energy (in J) of the emitted photoelectron is ($1 eV = 1.6 \times 10^{-19}$ J, $R_H = 13.6$ eV):
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$KE = E_{photon} - |E_{n}|$.
Updated On: Oct 14, 2026
- $3.02 \times 10^{-19}$
- $1.39 \times 10^{-18}$
- $1.93 \times 10^{-18}$
- $2.15 \times 10^{-19}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Photoelectric effect in atoms.
Step 2: Meaning
$KE = E_{photon} + E_{initial_state} = E_{photon} - |E_{initial_state}|$.
Step 3: Analysis
First excited state ($n=2$), $E_2 = -13.6/2^2 = -3.4$ eV. Absorbed photon energy $= 12.09$ eV. $KE = 12.09 - 3.4 = 8.69$ eV. Wait, checking ionization. To eject an electron, it needs to reach $n=\infty$ ($0$ eV). Energy required to ionize from $n=2$ is $3.4$ eV. $KE = 12.09 - 3.4 = 8.69$ eV. $8.69 \times 1.6 \times 10^{-19} = 13.9 \times 10^{-19} = 1.39 \times 10^{-18}$ J.
Step 4: Conclusion
The kinetic energy is $1.39 \times 10^{-18}$ J.
Final Answer: (B)
Photoelectric effect in atoms.
Step 2: Meaning
$KE = E_{photon} + E_{initial_state} = E_{photon} - |E_{initial_state}|$.
Step 3: Analysis
First excited state ($n=2$), $E_2 = -13.6/2^2 = -3.4$ eV. Absorbed photon energy $= 12.09$ eV. $KE = 12.09 - 3.4 = 8.69$ eV. Wait, checking ionization. To eject an electron, it needs to reach $n=\infty$ ($0$ eV). Energy required to ionize from $n=2$ is $3.4$ eV. $KE = 12.09 - 3.4 = 8.69$ eV. $8.69 \times 1.6 \times 10^{-19} = 13.9 \times 10^{-19} = 1.39 \times 10^{-18}$ J.
Step 4: Conclusion
The kinetic energy is $1.39 \times 10^{-18}$ J.
Final Answer: (B)
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