The probability of the particle in 1-D box is given by
\[ P = \left( \sqrt{\frac{{2}}{L}} \right)^2 \int_{0}^{L/4} \sin^2 \left( \frac{2\pi x}{L} \right) dx \]
\[ P = \frac{2}{2L} \int_{0}^{L/4} \left( 1 - \cos \left( \frac{4\pi x}{L} \right) \right) dx \]
Here \( n = 2 \) is the first excited state.
\[ P = \frac{1}{L} \left[ x \Big|_0^{L/4} - 0 \right] \]
\[ P = \frac{1}{4} \]
The energy of the particle in 1-D box is given by
\[ E = \frac{n^2 h^2}{8mL^2} \]
\[ E = \frac{4h^2}{8mL^2} = \frac{h^2}{2mL^2} \]
The energy of a particle in 1-D box is given by
\[ E = \frac{n^2 h^2}{8mL^2} \]
The least energy (\(n=1\)) can written as
\[ E = \frac{h^2}{8mL^2} \]
\[ E = \frac{(6.63 \times 10^{-34})^2}{8 \times 9.11 \times 10^{-31} \times (10^{-20})^2} \]
\[ E = \frac{(6.63 \times 10^{-31})^2}{8 \times 9.11 \times 10^{-31} \times 10^{-20}} \]
\[ E = 0.603 \times 10^{-17} \]
Converting it into eV
\[ E = 0.603 \times 10^{-17} \times \frac{1.6 \times 10^{-19}}{eV} \]
\[ E = 37 eV \]
Given:
The length of the box \( L = 1 \, \text{Å} \)
Mass of the neutron \( m = 1.67 \times 10^{-27} \, \text{kg} \)
We have to find the minimum energy \( E_1 \) (\( n = 1 \)).
The energy of the particle in 1-D box is given by
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
\[ E_1 = \frac{h^2}{8mL^2} \]
\[ E_1 = \frac{(6.625 \times 10^{-34})^2}{8 \times 1.67 \times 10^{-27} \times (10^{-10})^2} \]
\[ E_1 = 3.28 \times 10^{-21} \, \text{J} \]
\[ E_1 = \frac{3.28 \times 10^{-21}}{1.6 \times 10^{-19}} \, \text{eV} \]
\[ E_1 = 0.02 \, \text{eV} \]
The energy of a particle in the \(n\)-th state in a 1-D box is equal to the ground state energy multiplied by the square of the principal quantum number \(n\).
\[ E_n = n^2 E_1 \]Given:
n=1+5 =6 \[ E_6 = 6^2\times 18 eV \] \[ E_6 = 36\times 18 eV \] \[ E_6 = 648 eV \]