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– Thomas A. Edison

Questions on** **Atomic Physics and Quantum Effects with solution were posted earlier on this site. You can access all the posts by clicking on the label ‘atomic physics’ below this post or by trying a search using the search box provided on this page. Today we will discuss a few more practice questions (MCQ) in this section:

**(1)** The adjoining figure shows the energy level diagram of a hypothetical atom. A gaseous sample containing these atoms in the *ground state* is irradiated with photons in the band of energies from 7 eV to 9 eV. Consequently the excited atoms in the gas will emit photons of the following energies:

(a) All energies in the band from 7 eV to 9 eV

(b) Energies 6 eV, 10 eV and 15 eV

(c) Infinite number of discreet energies from 0 eV to 10 eV

(d) Energies 4 eV, 5 eV and 9 eV

(e) Infinite number of discreet energies from 0 eV to 15 eV

The atoms in the ground state (of energy – 15 eV) can absorb 9 eV from the incident photons and can get excited to the state of energy – 6 eV since it is an allowed state for the atom as per the energy level diagram given in the question. Once in the excited state of energy – 6 eV, the atom can undergo transition to the lower states of energy – 10 eV and – 15 eV in three ways:

(i) Jump first from – 6 eV level to – 10 eV level, emitting a 4 eV photon.

(ii) Jump next from – 10 eV level to – 15 eV level, emitting a 5 eV photon.

(iii) Jump directly from – 6 eV level to – 15 eV level, emitting a 9 eV photon.

Therefore the excited gas will emit photons of energies 4 eV, 5 eV and 9 eV as given in option (d).

**(2)** If the gaseous sample of atoms in the above question were irradiated with photons of *two *discreet energies 2 eV and 12 eV, what will be the energies of the photons emitted by the gas?

(a) Energies 7 eV and 12 eV

(b) All energies in the band from 7 eV to 12 eV

(c) Energies 3 eV, 4 eV, 5 eV, 7 eV, 9 eV, and 12 eV

(d) Energies 3 eV, 4 eV and 5 eV

(e) Infinite number of discreet energies from 0 eV to 12 eV

Incident photons of energy 2 eV cannot be absorbed by the atoms since there is no energy level which is higher by 2 eV with respect to the ground level. But the incident photons of energy 12 eV can be absorbed by the atoms since there is the energy level of – 3 eV which is higher by 12 eV with respect to the ground level of energy – 15 eV.

After getting excited to the level of – 3 eV, the atom can go to the ground state in *six* different ways:

(i) Jump first from – 3 eV level to – 6 eV level, emitting a 3 eV photon.

(ii) Jump next from – 6 eV level to – 10 eV level, emitting a 4 eV photon.

(iii) Jump next from – 10 eV level to – 15 eV level, emitting a 5 eV photon

(iv) The atom (actually the electron in the atom) can jump from – 3 eV level to – 10 eV level, emitting a 7 eV photon.

(v) The atom, after reaching – 6 eV level [as given in (i)] can jump directly to – 15 eV level, emitting a 9 eV photon.

(vi) The atom can jump directly from – 3 eV level to – 15 eV level, emitting a 12 eV photon.

Therefore the excited gas will emit photons of energies 3 eV, 4 eV, 5 eV, 7 eV, 9 eV, and 12 eV as given in option (c).

**(3) **The wave property of particles is evident in

(a) scattering of photons by electrons at rest

(b) Coolidge X-ray tube

(c) electron microscope

(d) scattering of α-particles by atomic nuclei

(e) photo electric cells

The correct option is (c). In an electron microscope electron beam is used (similar to light beam in optical microscope) because of the wave property of electrons. The wave length of electron at a few hundred volts of accelerating voltage is of the order of X-ray wave length so that much greater resolution and magnification compared to optical microscope is possible.

**(4) **In a hydrogen atom in the ground state suppose the orbital speed of the electron is ‘v’. If the mass of the electron is ‘m’ and the magnitude of its charge is ‘e’, what is the radius of the orbit of the electron?

(a) e^{2}/4πε_{0}mv^{2}

(b) e^{2}/4πε_{0}mv

(c) e/4πε_{0}mv^{2}

(d) mv^{2}/4πε_{0}e^{4}

(e) mv^{2}/4πε_{0}e^{2}

The centripetal force required for the circular motion of the electron is supplied by the electrostatic attractive force between the electron (of charge – e) and the nucleus (which is a proton of charge + e). Therefore we have

mv^{2}/r = (1/4πε_{0}) (e^{2}/r^{2}) where ε_{0} is the permittivity of free space.

This gives r = e^{2}/4πε_{0}mv^{2}

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