Understanding Electron Transitions in Hydrogen Atom
This problem involves understanding the energy levels and angular momentum of an electron within a hydrogen atom, specifically when it absorbs energy.
Key Concepts
- Energy Levels: In the Bohr model of the hydrogen atom, the energy of an electron is quantized and depends on the principal quantum number, denoted by '$n$'. The formula is given by: $E_n = -\frac{13.6}{n^2} \text{ eV}$ Here, '$n$' can be 1, 2, 3, ... corresponding to the ground state, first excited state, second excited state, and so on.
- Ground State: The ground state is the lowest energy level, which for hydrogen corresponds to '$n=1$'.
- Angular Momentum: The angular momentum ('$L$') of an electron in a hydrogen atom is also quantized and depends on '$n$'. The formula is: $L_n = n \frac{h}{2\pi}$ where '$h$' is Planck's constant.
- Energy Absorption: When an electron absorbs energy, it moves from a lower energy level to a higher energy level.
Step-by-Step Calculation
- Determine the initial state: The electron is initially in the ground state, so the initial principal quantum number is $n_i = 1$.
- Calculate the initial energy: Using the energy level formula for $n=1$: $E_1 = -\frac{13.6}{1^2} \text{ eV} = -13.6 \text{ eV}$
- Calculate the final energy: The electron absorbs 12.09 eV. The final energy ($E_f$) is the initial energy plus the absorbed energy: $E_f = E_1 + \Delta E$ $E_f = -13.6 \text{ eV} + 12.09 \text{ eV}$ $E_f = -1.51 \text{ eV}$
- Determine the final state (principal quantum number): We need to find the value of '$n$' for which the energy level is -1.51 eV. Using the energy level formula: $E_n = -\frac{13.6}{n^2}$ $-1.51 \text{ eV} = -\frac{13.6}{n^2}$ Rearranging the formula to solve for $n^2$: $n^2 = \frac{-13.6}{-1.51}$ $n^2 \approx 9$ Taking the square root: $n = \sqrt{9} = 3$ So, the electron transitions to the $n_f = 3$ energy level.
- Calculate the initial angular momentum: For the ground state ($n_i = 1$): $L_i = n_i \frac{h}{2\pi} = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi}$
- Calculate the final angular momentum: For the final state ($n_f = 3$): $L_f = n_f \frac{h}{2\pi} = 3 \times \frac{h}{2\pi} = 3\frac{h}{2\pi}$
- Calculate the increase in angular momentum: The increase is the difference between the final and initial angular momentum: $\Delta L = L_f - L_i$ $\Delta L = 3\frac{h}{2\pi} - \frac{h}{2\pi}$ $\Delta L = (3 - 1)\frac{h}{2\pi}$ $\Delta L = 2\frac{h}{2\pi}$
Conclusion
The increase in the electron's angular momentum is $2(h/2\pi)$. This corresponds to the transition from the $n=1$ state to the $n=3$ state.
Important Questions from Atoms
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