The "Particle in a 1D Box" is the simplest quantum mechanical system, illustrating how confinement leads to energy quantization. We consider a particle of mass $m$ free to move along the x-axis between $x=0$ and $x=L$, bounded by impenetrable walls.
The potential energy is defined as:
$$ V(x) = \begin{cases} 0 & \text{for } 0 < x < L \\ \infty & \text{for } x \le 0 \text{ and } x \ge L \end{cases} $$Because the potential is infinite outside the box, the probability of finding the particle there is zero. Thus, the wavefunction must be zero at the boundaries: $\psi(0) = 0$ and $\psi(L) = 0$.
Inside the box ($V=0$), the Time-Independent Schrödinger Equation is:
$$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \implies \frac{d^2\psi}{dx^2} + k^2\psi = 0 $$where $k = \frac{\sqrt{2mE}}{\hbar}$. The general solution to this differential equation is:
$$ \psi(x) = A\sin(kx) + B\cos(kx) $$1. At $x = 0$, $\psi(0) = 0$:
$$ A\sin(0) + B\cos(0) = 0 \implies B = 0 $$So, $\psi(x) = A\sin(kx)$.
2. At $x = L$, $\psi(L) = 0$:
$$ A\sin(kL) = 0 $$Since $A=0$ gives a trivial solution (no particle), we must have $\sin(kL) = 0$. This implies:
$$ kL = n\pi \implies k = \frac{n\pi}{L} $$where $n = 1, 2, 3, \dots$ (Notice $n=0$ is not allowed because it would mean $\psi(x)=0$ everywhere).
Substituting $k$ back into the definition of $E$:
$$ E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2 = \frac{n^2 \pi^2 \hbar^2}{2mL^2} $$Or in terms of $h$ ($\hbar = h/2\pi$):
$$ E_n = \frac{n^2 h^2}{8mL^2} $$To find the constant $A$, we normalize the wavefunction over the box:
$$ \int_0^L |\psi(x)|^2 dx = 1 \implies A^2 \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = 1 $$Using $\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}$, the integral evaluates to $L/2$. Therefore:
$$ A^2 \left(\frac{L}{2}\right) = 1 \implies A = \sqrt{\frac{2}{L}} $$The normalized wavefunctions are: $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$.
Given: $L = 1 \text{ nm} = 10^{-9} \text{ m}$, $m_e = 9.11 \times 10^{-31} \text{ kg}$, $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$.
Ground state ($n=1$): $E_1 = \frac{h^2}{8mL^2}$
First excited state ($n=2$): $E_2 = \frac{4h^2}{8mL^2} = 4E_1$
Energy difference: $\Delta E = E_2 - E_1 = 3E_1 = 3 \times \frac{(6.626 \times 10^{-34})^2}{8(9.11 \times 10^{-31})(10^{-9})^2}$
$\Delta E = 3 \times \frac{43.9 \times 10^{-68}}{72.88 \times 10^{-49}} = 3 \times 6.02 \times 10^{-20} \text{ J} = 1.806 \times 10^{-19} \text{ J}$
Convert to eV (divide by $1.6 \times 10^{-19}$):
$$ \Delta E = \frac{1.806 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.13 \text{ eV} $$For a box centered at the origin, the ground state is an even function: $\psi_1(x) = \sqrt{\frac{2}{L}} \cos\left(\frac{\pi x}{L}\right)$.
Probability $P = \int_0^{L/4} |\psi_1|^2 dx = \frac{2}{L} \int_0^{L/4} \cos^2\left(\frac{\pi x}{L}\right) dx$
Using $\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}$:
$P = \frac{1}{L} \int_0^{L/4} \left(1 + \cos\left(\frac{2\pi x}{L}\right)\right) dx = \frac{1}{L} \left[ x + \frac{L}{2\pi}\sin\left(\frac{2\pi x}{L}\right) \right]_0^{L/4}$
$P = \frac{1}{L} \left( \frac{L}{4} + \frac{L}{2\pi}\sin\left(\frac{\pi}{2}\right) \right) = \frac{1}{4} + \frac{1}{2\pi}$
$P \approx 0.25 + 0.159 = 0.409$ or $40.9\%$