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, \(m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times\) \(10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}\);
\(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}\)
- A 3
- B 5
- C 7
- D 9
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, \(\lambda=7 \mathrm{fm}\)
\(\begin{array}{c}
\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\
\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}
\end{array}\)
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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