Ncert Exercises & Solutions

Question 11.37:

Answer the following questions:

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (-1/3)e]. Why do they not show up in Millikan's oil-drop experiment?

(b) What is So Special about the combination e/m, Why do we not simply talk of e and m separately?

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic, Why is there an energy distribution of photoelectrons?

(e)The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: $E = hv, p = \frac{h}{λ}$

But while the value of $λ$ is physically significant, the value of v (and therefore, the value of the phase speed v $λ$ ) has no physical significance. Why?

(a)
Hint: Isolated quarks do not exist in nature.
Step 1: Find the force binding the quarks.
Quarks inside the protons and neutrons are bonded with strong nuclear force, which increases as the distance between them increases. So it's hard to separate quarks from the proton or neutron. Isolated quarks do not exist in nature.

(b)
Hint: Recall the dynamics of an electron.
Step 1: Find the basic relations between the electric field and the magnetic field.
The basic relations between the electric field and magnetic field are

(eV = \frac{1}{2} {mv}^{2}) and (eBv = \frac{{mv}^{2}}{r}) respectively.

These relations include e (electric charge), v (velocity), m (mass), V r(radius), and B (magnetic field). These relations give the value of the velocity of an electron as

(v = \sqrt{2 V (\frac{e}{m})}) and (v = Br (\frac{e}{m}))

respectively.

It can be observed from these relations that the dynamics of an electron are determined not by e and m separately, but by the ratio e/m.

(c)
Hint: At ordinary pressures, positive ions and electrons produced by the gas do not reach the electrodes.
Step 1: Find the behaviors of gas molecules at atmospheric pressure.
At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

(d)
Hint: Recall Einstein's photoelectric effect.
Step 1: Find the energy of surface electrons and electrons inside the metal.
The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.

(e)
Hint: $E=\frac{h^{2}}{2 \lambda^{2} m}$
Step 1: Find the momentum of matter waves.
According to de Broglie's hypothesis
$\lambda=\frac{h}{p}\\ \Rightarrow p=\frac{h}{\lambda}$

Step 2: Find the energy of a photon.
The energy of a photon is given by:
$E= hv=\frac{hc}{\lambda}$
Step 3: Find the energy of the moving particle.
$E=\frac{p^{2}}{2 m}=\frac{(h / \lambda)^{2}}{2 m}=\frac{h^{2}}{2 \lambda^{2} m}$
From the above relation, we can see that v has no direct significance on the relation of E and P.

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