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NCERT · Class 12 · Physics

Chapter 11: Dual Nature of Radiation and Matter

Practice each question, then open the answer or worked solution.

Questions

32 questions
Q11.36Written answer

Compute the typical de Broglie wavelength of an electron in metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This
indistinguishability has many fundamental implications, which you will explore in more advanced Physics courses.]

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Worked solution

Temperature, T = 27°C = 27 + 273 = 300 K

Mean separation between two electrons, r = 2 × 10−10 m

De Broglie wavelength of an electron is

\(\begin{array}{l}\lambda =\frac{h}{\sqrt{3mkT}}\end{array} \)

h = Planck’s constant = 6.6 × 10−34 Js

m = Mass of an electron = 9.11 × 10−31 kg

k = Boltzmann constant = 1.38 × 10−23 J/mol/K.

\(\begin{array}{l}\lambda =\frac{6.6\times 10^{-34}}{\sqrt{3\times 9.11\times 10^{-31}\times 1.38\times 10^{-23}\times 300}}\end{array} \)

≈ 6.2 x 10-9 m

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

Q11.37Written answer

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?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(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 = h ν, p = h/λ
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?

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Worked solution

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. This is because nuclear forces grow stronger if they are pulled apart. Therefore, it seems that fractional charges may exist in nature. The observable charges are an integral multiple of electrical charges (e).

(b) The relation between the electric field and the magnetic field,

eV = (1/2) mv2  and eBv = mv2/r

Here,

e = electric charge

v = velocity

V = potential

r = Radius

B = magnetic field

From these equations, it can be understood that the dynamics of an electron can be determined only by the ratio e/m and not by e and m separately.

(c)  At the atmospheric pressure, the ions in the gas do not have a chance of reaching their respective electrons due to collision and recombination with other molecules in the gas. At low pressures, ions have a chance to reach their respective electrons, which results in the flow of current.

(d)  The minimum energy required for an electron in the conduction band to get out of the metal is called the work function. These electrons occupy different energy levels, because of which, for the same incident radiation, electrons come out with different energies.

(e) The absolute value of the energy of a particle is arbitrary within the addictive constant. Therefore, the wavelength(λ) is significant, but the frequency (ν) of the electron does not have direct physical significance. Therefore, product νλ has no physical significance.