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Problem 1. Calculate de Broglie wavlength of an electron moving with velocity 2×104 m/s. [Given: mass of electron = 9.1 × 10-31 kg and Planck’s constant = 6.6 × 10-34Js].

(Amity Univer., May 2005) 

Problem 3. Find group velocity in terms of energy and momentum.


Problem 4. If the phase velocity of a ripple is by the relation vp = c1 + c2λ, where c1 and c2 are constant. Find the group velocity of the ripple.


Problem 5. The phase velocity of a soap bubble is 1/2. Find the group velocity of soap bubble. Hint: Problem 6. Use the uncertainty principle to deduce the lowest nature of energy of a particle of mass enclosed in a box of size ‘a’. Solution: The uncertainty relation is Problem 7. An electron is localised in a region of size (i) 1 cm (ii) 108 cm. Calculate order of kinetic energy of the electron in each case. Hint: Problem 8. Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the uncertainty in its velocity is equal to its velocity.

(Δx = λ, Δ = v).

Solution: From uncertainty principle Δx Δp = h Here Δx = λ Uncertainty is the measurement of the momentum

Problem 9. Show that the uncertainty principle is not observed in daily life, while it is valid for microscopic world.

Solution: Case I: Uncertainty principle for macroscopic world (daily life). Let us consider a sand particle of mass m = 106 kg and its size is 10-4 m. Therefore uncertainty in the measurement of the position of the sand particle

This gives the uncertainty in the velocity of the sand particle which is too small. Therefore uncertainty principle is not effective in daily life. Case II. Uncertainty principle in microscopic world (subatomic world): Since the matter consists of atoms and molecules and atom comprises electrons. The mass of electron is me = 9.1 × 1031 kg The size of electron = lA = 1010 m From uncertainty principle Uncertainty in the velocity of electron

= 0.74 × 107 m/s = 7.4 × 106 m/sec C.

which is observable. Thus this uncertainty in velocity of the microscopic produces unheard result. Hence uncertainty principle is effective in subatomic world.
Problem 10. Find the de Broglie wavelength of a 0.01 kg pellet having a velocity of 10 m/s.

Hint:  Problem 11. What is the uncertainty in the location of a photon of wavelength 3000 A if this wavelength is known to an accuracy of one part in million?

Hint: Problem 12. What is de Broglie wavelength of an electron which has been accelerated from rest through a potential difference of 100 V.


Problem 13. Calculate de Broglie wavelength associated with electron having 10 keV kinetic energy, where me = 9.1 × 10-31 kg and h = 6.6 × 10-34 Js.

Hint: Problem 14. An electron of energy 200 eV is passed through a circular hole of radius 106 m. What is the uncertainty introduced in the angle of emergence?

Hint: Problem 15. The average period of that elapses between excitation of an atom and the time it emits radiation is 10-8 sec. Find the uncertainty in the frequency of light emitted.

Hint: Problem 16. Find the uncertainty in the kinetic energy of a proton confined to a nucleus size of 10-14 m. (mass of proton = 1.66 × 1027 kg).


Problem 17. Which of the wave functions shown in Figure 9.13 have no physical significance for values of x as shown. Give reasons:

Fig. 9.13 (a)

Fig. 9.13 (b)

Fig. 9.13 (c)


Fig. 9.13 (d)

Fig. 9.13 (e)

Fig. 9.13 (f)


(b) Multivalued

(c) Discontinuous and becomes infinite

(d) Discontinuous

(f) becomes infinite

Solution: (i) Since the above function must obey the normalisation condition, we must have


Problem 19.The minimum energy of a particles trapped in a one dimensional potential well is 4×10-18 J. What are the next three energies?