When a particle is restricted to move aong `x`axis between `x =0` and `x = a`, where a is of nanometer 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 realated to the linear momentum `p` of the particle according to the de Breogile relation. The energy of the particl e of mass m is reelated 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 loop in the standing wave.
Use the model decribed above to answer the following three questions for a particle moving in the line `x = 0` to `x =a`. Take `h = 6.6 xx 10^(-34) J s` and `e = 1.6 xx 10^(-19) C`.
If the mass of the particle is `m = 1.0 xx 10^(-30) kg` and `a = 6.6 nm`, the energy of the particle in its ground state is closet to

Options

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

Answer

B

Explanation

`E=h^(2)/(2m4a^(2))implies ((6.6xx10^(-34))^(2))/(2(1.0xx10^(-30))xx4xx(6.6xx10^(-9))^(2)) 1/e`
`E=8 me V`