The following three questions [(i), (ii) and (iii)] involving de Broglie’s matter waves were included under Linked Comprehension Type Multiple Choice Questions (single answer type) in the IIT-JEE 2009 question paper (Paper I):
Paragraph for Questions (i), (ii) and (iii)
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 = p2/2m
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 × 10–34 J s and e = 1.6 × 10–19 C.
Question i
The allowed energy for the particle for a particular value of n is proportional to
(A) a–2
(B) a–3/2
C) a–1
(D) a2.
If there are n loops in the standing wave, we have
a = n λ/2 from which λ = 2a/n
[Note that the distance between consecutive nodes is λ/2]
From de Broglie relation, momentum p = h/λ = nh/2a on substituting for λ.
Now, energy E = p2/2m = n2h2/8a2m
Therefore, E α a–2 [Option (A)].
Question ii
If the mass of the particle is m = 1.0×10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to
(A) 0.8 meV
(B) 8 meV
(C) 80 meV
(D) 800 meV
In the ground state, n = 1 so that energy E = h2/8a2m
Therefore, E = (6.6 × 10–34)2/[8×(6.6×10–9)2×10–30]
=10–20/8 joule = 10–20/(8×1.6×10–19) electron volt
= 10–1/(8×1.6) eV = 0.0078 eV, nearly = 7.8 meV (milli electron volt)
The correct option (nearest to the answer) is (B).
[Note that the symbol for million electron volt is MeV]
Question iii
The speed of the particle, that can take discrete values, is proportional to
(A) n–3/2
(B) n–1
(C) n1/2
(D) n
The energy of the particle is given by E = p2/2m = n2h2/8a2m
Therefore, ½ mv2 = n2h2/8a2m
From this the speed, v α n [Option (D)]
