To determine the energy of an electron in the first excited state (\( n = 2 \)) within a one-dimensional (1D) box of length \( L = 1 \, \text{Å} \), we use the quantum mechanical model for a particle in a box. The energy levels are quantized and given by the formula: \[ E_n = \frac{n^2 h^2}{8mL^2} \] where:
- \( E_n \): energy of the electron in the \( n^{\text{th}} \) state
- \( n \): principal quantum number (for the first excited state, \( n = 2 \))
- \( h \): Planck's constant (\( 6.626 \times 10^{-34} \, \text{J·s} \))
- \( m \): mass of the electron (\( 9.109 \times 10^{-31} \, \text{kg} \))
- \( L \): length of the box (\( 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \))
- Substitute the known values into the equation:
\[ E_2 = \frac{(2)^2 \times (6.626 \times 10^{-34} \, \text{J·s})^2}{8 \times (9.109 \times 10^{-31} \, \text{kg}) \times (1 \times 10^{-10} \, \text{m})^2} \] - Calculate the numerator:
\[ (2)^2 = 4 \] \[ (6.626 \times 10^{-34})^2 = 4.392 \times 10^{-67} \, \text{J}^2 \cdot \text{s}^2 \] \[ \text{Numerator} = 4 \times 4.392 \times 10^{-67} = 1.757 \times 10^{-66} \, \text{J}^2 \cdot \text{s}^2 \] - Calculate the denominator:
\[ 8 \times 9.109 \times 10^{-31} \times (1 \times 10^{-10})^2 = 8 \times 9.109 \times 10^{-31} \times 1 \times 10^{-20} \] \[ = 7.287 \times 10^{-50} \, \text{kg·m}^2 \] - Compute the energy \( E_2 \):
\[ E_2 = \frac{1.757 \times 10^{-66}}{7.287 \times 10^{-50}} = 2.414 \times 10^{-17} \, \text{J} \] - Convert energy from joules to electron volts (1 eV = \( 1.602 \times 10^{-19} \, \text{J} \)):
\[ E_2 = \frac{2.414 \times 10^{-17} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 150.6 \, \text{eV} \] - Rounding to an appropriate number of significant figures:
\[ E_2 \approx 150.4 \, \text{eV} \]
Therefore, the energy of the electron in the first excited state is approximately 150.4 eV.