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 = p*^{2}/2*m*

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) *a*^{2.}

If there are* n* loops in the standing wave, we have

*a = n* *λ/*2 from which *λ = *2*a/n*

[Note that the distance between consecutive nodes is *λ/*2]

From de Broglie relation, momentum *p = h/λ* = *nh/*2*a* on substituting for *λ*.

Now, energy *E = p*^{2}/2*m* = *n*^{2}*h*^{2}*/*8*a*^{2}*m*^{}

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 *= *h*^{2}*/*8*a*^{2}*m*

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) *n*^{1/2}

(D)* n*

The energy of the particle is given by *E = p*^{2}/2*m* = *n*^{2}*h*^{2}*/*8*a*^{2}*m*

Therefore, ½ *mv*^{2} = *n*^{2}*h*^{2}*/*8*a*^{2}*m*

From this the speed, *v*** α n **[Option (D)]

**Linked Comprehension Type Multiple Choice Questions on nuclear physics (with solution) here**

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