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JEE Advanced · Physics · 26. Dual Nature

A proton is fired from very far away towards a nucleus with charge \(Q=120 e\), where \(e\) is the electronic charge. It makes a closest approach of \(10 \mathrm{fm}\) to the nucleus. The de Broglie wavelength (in units of \(\mathrm{fm}\) ) of the proton at its start is: (take the proton mass, \(m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times\) \(10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}\);

\(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}\)

  1. A 3
  2. B 5
  3. C 7
  4. D 9
Verified Solution

Answer & Solution

Correct Answer

(C) 7

Step-by-step Solution

Detailed explanation

From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system

\(\begin{array}{c}

\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\

\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}

\end{array}\)

Putting the values of \(4 \pi \varepsilon_{0}, r, h, q_{1}, q_{2}\) and \(m\) we get, deBroglie wavelength of proton, \(\lambda=7 \mathrm{fm}\)
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