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Dual Nature of Radiation question

2009 · Shift 1 · Q54
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Dual Nature of Radiation question

2009 · Shift 1 · Q54

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.If the mass of the particle is m=1.0×10−30m=1.0\times10^{-30}m=1.0×10−30 kg and a=6.6a=6.6a=6.6 nm, the energy of the particle in its ground state is closest to
  1. A
    0.8 meV
  2. B
    8 meV
  3. C
    80 meV
  4. D
    800 meV
View written solutionFree

Correct answer: B

Step-by-step Derivation

  1. Model the System: The problem describes a particle of mass m confined to a one-dimensional box of length a. This is a classic quantum mechanics problem known as the "particle in a box". The allowed states of the particle are described by standing waves.

  2. Standing Wave Condition: For a standing wave to form in a region of length a with nodes at both ends (x=0 and x=a), the length a must be an integer multiple of half-wavelengths (λ/2\lambda/2λ/2). a=nλ2a = n \frac{\lambda}{2}a=n2λ​ where n is a positive integer (n=1,2,3,...n = 1, 2, 3, ...n=1,2,3,...). The quantum number n corresponds to the number of loops in the standing wave. From this, we can express the allowed wavelengths: λn=2an\lambda_n = \frac{2a}{n}λn​=n2a​

  3. De Broglie Relation: The de Broglie hypothesis relates a particle's momentum p to its wavelength λ\lambdaλ: p=hλp = \frac{h}{\lambda}p=λh​ Substituting the allowed wavelengths, we get the quantized momentum: pn=hλn=h2a/n=nh2ap_n = \frac{h}{\lambda_n} = \frac{h}{2a/n} = \frac{nh}{2a}pn​=λn​h​=2a/nh​=2anh​

  4. Energy Quantization: The energy of the particle is given as kinetic energy, related to its momentum: E=p22mE = \frac{p^2}{2m}E=2mp2​ Substituting the expression for quantized momentum pnp_npn​, we get the quantized energy levels EnE_nEn​: En=pn22m=(nh/2a)22m=n2h28ma2E_n = \frac{p_n^2}{2m} = \frac{(nh/2a)^2}{2m} = \frac{n^2 h^2}{8ma^2}En​=2mpn2​​=2m(nh/2a)2​=8ma2n2h2​

  5. Calculate Ground State Energy: The ground state is the lowest energy state, which corresponds to n = 1. E1=(1)2h28ma2=h28ma2E_1 = \frac{(1)^2 h^2}{8ma^2} = \frac{h^2}{8ma^2}E1​=8ma2(1)2h2​=8ma2h2​

  6. Substitute Given Values:

    • Planck's constant, h=6.6×10−34h = 6.6 \times 10^{-34}h=6.6×10−34 J-s
    • Mass of the particle, m=1.0×10−30m = 1.0 \times 10^{-30}m=1.0×10−30 kg
    • Length of the box, a=6.6a = 6.6a=6.6 nm =6.6×10−9= 6.6 \times 10^{-9}=6.6×10−9 m

    E1=(6.6×10−34)28×(1.0×10−30)×(6.6×10−9)2E_1 = \frac{(6.6 \times 10^{-34})^2}{8 \times (1.0 \times 10^{-30}) \times (6.6 \times 10^{-9})^2}E1​=8×(1.0×10−30)×(6.6×10−9)2(6.6×10−34)2​

  7. Calculate the Energy in Joules (J): E1=(6.6)2×10−688×10−30×(6.6)2×10−18E_1 = \frac{(6.6)^2 \times 10^{-68}}{8 \times 10^{-30} \times (6.6)^2 \times 10^{-18}}E1​=8×10−30×(6.6)2×10−18(6.6)2×10−68​ The (6.6)2(6.6)^2(6.6)2 terms cancel out, simplifying the calculation: E1=10−688×10−30×10−18=10−688×10−48E_1 = \frac{10^{-68}}{8 \times 10^{-30} \times 10^{-18}} = \frac{10^{-68}}{8 \times 10^{-48}}E1​=8×10−30×10−1810−68​=8×10−4810−68​ E1=18×10−68+48=0.125×10−20 JE_1 = \frac{1}{8} \times 10^{-68+48} = 0.125 \times 10^{-20} \text{ J}E1​=81​×10−68+48=0.125×10−20 J E1=1.25×10−21 JE_1 = 1.25 \times 10^{-21} \text{ J}E1​=1.25×10−21 J

  8. Convert Energy to milli-electron Volts (meV): First, convert from Joules (J) to electron-volts (eV) using the conversion factor 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}1 eV=1.6×10−19 J. E1(in eV)=1.25×10−21 J1.6×10−19 J/eV=1.251.6×10−2 eVE_1 (\text{in eV}) = \frac{1.25 \times 10^{-21} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = \frac{1.25}{1.6} \times 10^{-2} \text{ eV}E1​(in eV)=1.6×10−19 J/eV1.25×10−21 J​=1.61.25​×10−2 eV E1(in eV)=0.78125×10−2 eV=0.0078125 eVE_1 (\text{in eV}) = 0.78125 \times 10^{-2} \text{ eV} = 0.0078125 \text{ eV}E1​(in eV)=0.78125×10−2 eV=0.0078125 eV Next, convert from eV to meV (1 eV=1000 meV1 \text{ eV} = 1000 \text{ meV}1 eV=1000 meV). E1(in meV)=0.0078125 eV×1000 meV/eV=7.8125 meVE_1 (\text{in meV}) = 0.0078125 \text{ eV} \times 1000 \text{ meV/eV} = 7.8125 \text{ meV}E1​(in meV)=0.0078125 eV×1000 meV/eV=7.8125 meV

  9. Compare with Options: The calculated ground state energy is approximately 7.81257.81257.8125 meV. This value is closest to option B.

    • A: 0.8 meV
    • B: 8 meV
    • C: 80 meV
    • D: 800 meV

    Therefore, the correct option is B.

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