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Dual Nature of Radiation question

2009 · Shift 1 · Q53
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Dual Nature of Radiation question

2009 · Shift 1 · Q53

JEE AdvancedPhysicsDual Nature of RadiationMCQ+3 / −1
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=p22mE = {{{p^2}} \over {2m}}E=2mp2​. 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=6.6×10−34h = 6.6 \times {10^{ - 34}}h=6.6×10−34 J-s and e=1.6×10−19e = 1.6 \times {10^{ - 19}}e=1.6×10−19 C.The allowed energy for the particle for a particular value of nnn is proportional to
  1. A
    a−2{a^{ - 2}}a−2
  2. B
    a−3/2{a^{ - 3/2}}a−3/2
  3. C
    a−1{a^{ - 1}}a−1
  4. D
    a2{a^2}a2
View written solutionFree

Correct answer: A

  1. Standing wave condition

For a particle confined between x=0x=0x=0 and x=ax=ax=a, the wave must have nodes at both ends. Hence the allowed wavelengths satisfy

a=n(λ2),n=1,2,3,…a = n\left(\frac{\lambda}{2}\right), \qquad n=1,2,3,\dotsa=n(2λ​),n=1,2,3,…

So,

λ=2an.\lambda = \frac{2a}{n}.λ=n2a​.

  1. Use de Broglie relation

The de Broglie relation is

p=hλ.p = \frac{h}{\lambda}.p=λh​.

Substituting λ=2an\lambda = \dfrac{2a}{n}λ=n2a​,

p=h2a/n=nh2a.p = \frac{h}{2a/n} = \frac{nh}{2a}.p=2a/nh​=2anh​.

  1. Energy of the particle

Given,

E=p22m.E = \frac{p^2}{2m}.E=2mp2​.

Substitute p=nh2ap = \dfrac{nh}{2a}p=2anh​:

= \frac{n^2 h^2}{8ma^2}.$$ 4. **Dependence on $a$** For a particular value of $n$, $m$, and $h$ are constants. Therefore, $$E \propto \frac{1}{a^2} = a^{-2}.$$ 5. **Option check** - A: $a^{-2}$ ✅ - B: $a^{-3/2}$ ❌ - C: $a^{-1}$ ❌ - D: $a^2$ ❌ Therefore, the correct option is **A**.
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