When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C.
Q. If the mass of the particle is `m=1.0xx10^(-30)`kg and `alpha=6.6nm`, the energy of the particle in its ground state is closest to

Options

A. 0.8 meV
B. 8 meV
C. 80 meV
D. 800 meV

Explanation

To find the energy of a particle in its ground state when it is confined to move between \( x = 0 \) and \( x = \alpha \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
The particle is confined in a one-dimensional box of length \( \alpha \). The allowed energies correspond to standing waves with nodes at the ends of the box.

2. **Using the de Broglie Wavelength:**
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