If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

When a particle is restricted to move along 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 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 $$E = {{{p^2}} \over {2m}}$$. 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 in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

The allowed energy for the particle for a particular value of $$n$$ is proportional to

When a particle is restricted to move along 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 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 $$E = {{{p^2}} \over {2m}}$$. 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 in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

If the mass of the particle is $$m=1.0\times10^{-30}$$ kg and $$a=6.6$$ nm, the energy of the particle in its ground state is closest to

When a particle is restricted to move along 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 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 $$E = {{{p^2}} \over {2m}}$$. 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 in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

The speed of the particle, that can take discrete values, is proportional to