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. $3.02 \times 10^{-19}$
• B. $1.39 \times 10^{-18}$
• C. $1.93 \times 10^{-18}$
• D. $2.15 \times 10^{-19}$

Hint:

$KE = E_{photon} - |E_{n}|$.

Answer 1

The Correct Option is A

Step 1: Understand the setup.
A hydrogen atom is sitting in its first excited state, which means the electron is in $n=2$. A photon of energy $12.09$ eV hits it and an electron is thrown out. We need the kinetic energy of that ejected electron.

Step 2: Energy of the starting level.
The energy of an electron in hydrogen is $E_n = -\dfrac{13.6}{n^2}$ eV. For $n=2$: \[ E_2 = -\frac{13.6}{4} = -3.4 \text{ eV} \] So the electron is bound by $3.4$ eV.

Step 3: How much energy is used to free it.
To pull the electron completely away (up to $n=\infty$ where energy is $0$), we must spend $3.4$ eV. The rest of the photon energy becomes the speed (kinetic energy) of the free electron.

Step 4: Find the kinetic energy in eV.
\[ KE = E_{photon} - |E_2| = 12.09 - 3.4 = 8.69 \text{ eV} \]

Step 5: Convert to joules.
Using $1$ eV $= 1.6 \times 10^{-19}$ J: \[ KE = 8.69 \times 1.6 \times 10^{-19} = 13.9 \times 10^{-19} = 1.39 \times 10^{-18} \text{ J} \]

Step 6: Conclusion.
So the kinetic energy of the emitted photoelectron is $1.39 \times 10^{-18}$ J. \[ \boxed{1.39 \times 10^{-18}\ \text{J}} \]