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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 toA. `0.8 me V`B. `8 meV`C. `80 meV`D. `800 meV` |
Answer» Correct Answer - B `E = (n^(2)h^(2))/(8 ma^(2))` For ground state `n = 1 rArr E_(1) = ((1)^(2)(6.6 xx 10^(-34))^(2))/(8 xx 10^(-30) xx (6.6 xx 10^(-9))^(2) xx 1.6 xx 10^(-19)) = 8 m eV` |
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