JEE Challenger
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PhyHardAdv 2012 • Paper 1

Initial de Broglie Wavelength of Incident Proton Approaching Charged Nucleus

A proton is projected from a large distance towards a nucleus of charge Q=120eQ = 120 e, where ee is the elementary charge. Its distance of closest approach to the nucleus turns out to be 10 fm10\text{ fm}. The initial de Broglie wavelength (expressed in fm\text{fm}) of the proton is: (Given mass of proton mp=(5/3)×10−27 kgm_p = (5/3) \times 10^{-27}\text{ kg}; h/e=4.2×10−15 J⋅s/Ch/e = 4.2 \times 10^{-15}\text{ J}\cdot\text{s/C}; 14πε0=9×109 m/F\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ m/F}; 1 fm=10−15 m1\text{ fm} = 10^{-15}\text{ m})

Official Numerical Answer7

Topics & Concepts

Dual Nature of Radiation and MatterDe Broglie Wavelength

Step-by-Step Solution

To find the initial de Broglie wavelength of the incident proton, we use the principle of conservation of mechanical energy and the formula for de Broglie wavelength.

1. Conservation of Mechanical Energy

When the proton is at a very large distance from the nucleus, its potential energy is zero, and its total energy is equal to its initial kinetic energy (KiK_i). Ki=12mpv02K_i = \frac{1}{2} m_p v_0^2

At the distance of closest approach (r0r_0), the proton momentarily comes to rest relative to the nucleus (assuming a head-on approach), so its kinetic energy becomes zero, and its energy is purely electrostatic potential energy (UU): U=14πε0qQr0U = \frac{1}{4\pi\varepsilon_0} \frac{q Q}{r_0}

Given:

  • q=eq = e (charge of proton)
  • Q=120eQ = 120 e (charge of nucleus)
  • r0=10 fm=10×10−15 m=10−14 mr_0 = 10 \text{ fm} = 10 \times 10^{-15} \text{ m} = 10^{-14} \text{ m}
  • k=14πε0=9×109 m/Fk = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ m/F}

Equating initial kinetic energy to potential energy at closest approach: Ki=ke(120e)r0=120ke2r0K_i = k \frac{e (120 e)}{r_0} = \frac{120 k e^2}{r_0}

2. Initial Momentum and de Broglie Wavelength

The initial momentum pp of the proton is related to its kinetic energy KiK_i by: p=2mpKi=2mp(120ke2r0)=e240mpkr0p = \sqrt{2 m_p K_i} = \sqrt{2 m_p \left( \frac{120 k e^2}{r_0} \right)} = e \sqrt{\frac{240 m_p k}{r_0}}

The initial de Broglie wavelength λ\lambda is given by: λ=hp=he240mpkr0=(he)1240mpkr0\lambda = \frac{h}{p} = \frac{h}{e \sqrt{\frac{240 m_p k}{r_0}}} = \left( \frac{h}{e} \right) \frac{1}{\sqrt{\frac{240 m_p k}{r_0}}}

3. Substitution of Numerical Values

Given values:

  • mp=53×10−27 kgm_p = \frac{5}{3} \times 10^{-27} \text{ kg}
  • he=4.2×10−15 J⋅s/C\frac{h}{e} = 4.2 \times 10^{-15} \text{ J}\cdot\text{s/C}
  • k=9×109 m/Fk = 9 \times 10^9 \text{ m/F}
  • r0=10−14 mr_0 = 10^{-14} \text{ m}

First, evaluate the term inside the square root: 240mpkr0=240×(53×10−27)×(9×109)10−14\frac{240 m_p k}{r_0} = \frac{240 \times \left(\frac{5}{3} \times 10^{-27}\right) \times \left(9 \times 10^9\right)}{10^{-14}} 240mpkr0=400×9×10−1810−14=3600×10−1810−14=3600×10−4=0.36\frac{240 m_p k}{r_0} = \frac{400 \times 9 \times 10^{-18}}{10^{-14}} = \frac{3600 \times 10^{-18}}{10^{-14}} = 3600 \times 10^{-4} = 0.36

Taking the square root: 240mpkr0=0.36=0.6\sqrt{\frac{240 m_p k}{r_0}} = \sqrt{0.36} = 0.6

Now, substitute back into the wavelength formula: λ=4.2×10−150.6=7×10−15 m=7 fm\lambda = \frac{4.2 \times 10^{-15}}{0.6} = 7 \times 10^{-15} \text{ m} = 7 \text{ fm}

Thus, the initial de Broglie wavelength of the proton expressed in fm\text{fm} is 77.

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