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When a particle is restricted to move along x-axis between x = 0 and x = 4, 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 r = 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 'r' 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 in the line x=0 to x=a. Take h=6.6 xx 10^(-34) Js 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 close to

Answer»

0.8meV
8meV
80meV
800meV

Solution :The mass of the particle in the box is `m=1.0 xx 10^(39) kg, a=6.6nm`
The energy of the particle in the ground state
`E_(n ML)=(l^(2).h^(2))/(8 ma^(2)E)` in electron volt units
`RARR` E in electron volts.
`=((6.6 xx 10^(-34))^(2))/(8 xx 1.0 xx 10^(-30) xx (6.6 xx 10^(-9))^(2) xx 1.6 xx 10^(-19))`
=7.8meV =8meV


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