Question

A proton is fired very far away towards a nucleus with charge $Q=120e$, where e is the electronic charge. It makes a closest approach of 10fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is?

& {{m}_{p}}=\dfrac{5}{3}\times {{10}^{-27}}kg\text{ and }\dfrac{h}{e}=4.2\times {{10}^{-15}}J.s/C \\\ & \dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}m/F \\\ & 1fm={{10}^{-15}}m \\\ \end{aligned}$$ A.7fm B.8fm C.9fm D.10fm

Answer 1

The closest approach of the proton is the indication of the kinetic energy possessed by the proton. The electrostatic energy or the electric field strength of the target nucleus will be equal to the kinetic energy of the proton at the closest approach.

Complete answer:
As we know the closest approach is the minimum distance between a nucleus and proton that can be attained during an experiment with protons of a specific velocity. The higher the velocity, higher will be the kinetic energy possessed by it.
The kinetic energy is given by:
$KE=\dfrac{1}{2}m{{v}^{2}}$
Now let us consider the nucleus of the target atom. The electric field strength is given by:
$E=\dfrac{Qq}{4\pi {{\varepsilon }_{0}}r}$,
where $Q=120e,q=e\text{ and r is the closest approach}$
We know that the closest approach happens when the kinetic energy of the proton equals the electrostatic energy of the nucleus. So we can equate them as –

& KE=E \\\ & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{Qq}{4\pi {{\varepsilon }_{0}}} \\\ & \text{and,}v=\sqrt{\dfrac{2Qq}{4\pi {{\varepsilon }_{0}}m}} \\\ \end{aligned}$$ So, now we have the velocity of the proton, let us find the de Broglie wavelength using this velocity, De Broglie wavelength $$\lambda $$ is given by – $$\begin{aligned} & \lambda =\dfrac{h}{mv} \\\ & \Rightarrow \lambda =\dfrac{h}{m\sqrt{\dfrac{2Qq}{4\pi {{\varepsilon }_{0}}rm}}} \\\ & \Rightarrow \lambda =h\sqrt{\dfrac{4\pi {{\varepsilon }_{0}}r}{2Qqm}} \\\ & \text{ = }h\sqrt{\dfrac{4\pi {{\varepsilon }_{0}}r}{240{{e}^{2}}m}} \\\ & \text{ =}\dfrac{h}{e}\sqrt{\dfrac{10\times {{10}^{-15}}}{240\times 9\times {{10}^{9}}\times \dfrac{5}{3}\times {{10}^{-27}}}} \\\ & \text{ =}4.2\times {{10}^{-15}}\sqrt{\dfrac{{{10}^{-14}}}{3600\times {{10}^{-18}}}} \\\ & \text{ =4}\text{.2}\times {{10}^{-15}}\times \dfrac{100}{60} \\\ & \text{ =7}\times \text{1}{{\text{0}}^{-15}}m \\\ & \therefore \lambda =7fm \\\ \end{aligned}$$ **The correct answer is option A.** **Additional Information:** In an experiment, the closest approach is identified when the proton rebounds when fired. This was Rutherford's scattering experiment based on. **Note:** The unit of measurement fermi meter is used in nuclear experiments. The distance of closest approach and the nuclear radius are measured in fermi meter scale, whereas the atomic radius is measured using the nano meters or angstrom.