Ground State Energy of Particle with Given Mass and Length

Based on the quantum mechanical model of a particle of mass $m$ confined to a one-dimensional box of length $a$ with rigid walls:

1. Quantization of Momentum:
   The standing wave condition for a particle trapped in a one-dimensional box of length $a$ requires that an integral number of half-wavelengths fit within the length:
   $n \left(\frac{\lambda}{2}\right) = a \implies \lambda = \frac{2a}{n}, \quad \text{where } n = 1, 2, 3, \dots$

Using the de Broglie relation $p = \frac{h}{\lambda}$, the allowed values of momentum are:
$p_n = \frac{nh}{2a}$

1. Energy Levels:
   The energy of the particle in state $n$ is purely kinetic and is given by:
   $E_n = \frac{p_n^2}{2m} = \frac{n^2 h^2}{8 m a^2}$

2. Ground State Energy ($n = 1$):
   For the ground state ($n = 1$):
   $E_1 = \frac{h^2}{8 m a^2}$

Given values:

• Mass of the particle, $m = 1.0 \times 10^{-30} \text{ kg}$
• Width of the box, $a = 6.6 \text{ nm} = 6.6 \times 10^{-9} \text{ m}$
• Planck's constant, $h \approx 6.6 \times 10^{-34} \text{ J}\cdot\text{s}$

Substitute these values into the energy equation:
$E_1 = \frac{(6.6 \times 10^{-34} \text{ J}\cdot\text{s})^2}{8 \times (1.0 \times 10^{-30} \text{ kg}) \times (6.6 \times 10^{-9} \text{ m})^2}$

$E_1 = \frac{6.6^2 \times 10^{-68}}{8 \times 1.0 \times 10^{-30} \times 6.6^2 \times 10^{-18}} \text{ J} = \frac{10^{-68}}{8 \times 10^{-48}} \text{ J} = \frac{1}{8} \times 10^{-20} \text{ J} = 1.25 \times 10^{-21} \text{ J}$

1. Conversion to meV:
   Converting the energy into electron-volts ($\text{eV}$), where $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$:
   $E_1 = \frac{1.25 \times 10^{-21}}{1.6 \times 10^{-19}} \text{ eV} \approx 7.81 \times 10^{-3} \text{ eV} = 7.81 \text{ meV} \approx 8 \text{ meV}$

Therefore, the ground state energy of the particle is closest to $8\text{ meV}$, which corresponds to Option (B).