## Wednesday, July 15, 2009

### IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions involving de Broglie Waves

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)]

You will find IIT-JEE 2009 Linked Comprehension Type Multiple Choice Questions on nuclear physics (with solution) here

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