When a particle is restricted to move along x- axis between `x = 0 and x = 4 ` whwre a is opf nanometer demension , its energy can take only certain spscfic values . The allowed energies of the particles only in such a restiricted regain , correspond to the formation of standing wave with nodes at its end ` x = 0 and x = a `.The wavelength of this standing wave is related to the linear momentum p of the paarticle according to the de Broglie relation .The energy of the particle of mass `m` is reated to its linear momentum as `E = (p^(2))/(2m)` . thus , the energy of the particle can be denoted by a quantum number `n` taking value `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 there question 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 energyof the particle in its ground state is closest to
A
`0.8 meV`
B
`8 meV`
C
`80 meV`
D
`800 meV`
Text Solution
Verified by Experts
The correct Answer is:
B
For ground state `n = 1`, Given `m = 1.0 xx 10^(-30)kg , a= 6.6 xx 10^(9) m ` `:. E = (t^(2) xx (6.6 xx 10^(-34)) ^(2))/(8 xx 1 xx 10^(-30) xx (6.6 xx 10^(-9))^(2)) 1 = 8 meV`
