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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Physics
Chapter
Dual Nature of Matter

JEE Advanced 2012 · Paper 1 · Physics · Dual Nature of Matter

A proton is fired from very far away towards a nucleus with charge...

A proton is fired from very far away towards a nucleus with charge Q=120eQ=120 e, where ee is the electronic charge. It makes a closest approach of 10fm10 \mathrm{fm} to the nucleus. The de Broglie wavelength (in units of fm\mathrm{fm} ) of the proton at its start is: (take the proton mass, mp=(5/3)×10−27 kg;h/e=4.2×m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times 10−15 J.S/C10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}; 14πε0=9×109 m/F;1fm=10−15 m\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}

Correct answer

7

Step-by-step solution

From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system 12mv2=14πε0q1q2r∴p22m=14π∈0q1q2r∴12m(h2λ2)=14π∈0q1q2r∴λ=4πϵ0r⋅h2q1q2(2m)\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 4πε0,r,h,q1,q24 \pi \varepsilon_{0}, r, h, q_{1}, q_{2} and mm we get, deBroglie wavelength of proton, λ=7fm\lambda=7 \mathrm{fm}
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