Question

Consider the radius of a nucleus to be \(10^{-15} \mathrm{~m} .\) If an electron is assumed to be in such nucleus, what ill be its energy? $\left(\mathrm{me}=9.1 \times 10^{-31} \mathrm{~kg} \cdot \mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(5.59 \times 10^{3} \mathrm{MeV}\) (B) \(9.55 \times 10^{3} \mathrm{MeV}\) (C) \(5.95 \times 10^{3} \mathrm{MeV}\) (D) \(7.45 \times 10^{3} \mathrm{MeV}\)

Short answer

The electron's energy, when assumed to be in the nucleus with a radius of \(10^{-15} \mathrm{m}\), is approximately \(5.95 \times 10^{3} \mathrm{MeV}\). Therefore, the correct answer is (C).

Step-by-step solution

Write down the Heisenberg Uncertainty Principle formula

Heisenberg Uncertainty Principle states that the product of the uncertainty in position, denoted by \(\Delta x\), and the uncertainty in momentum, denoted by \(\Delta p\), is greater than or equal to a constant involving Planck's constant h: \[\Delta x \Delta p \geq \frac{h}{4\pi}\]

Calculate the uncertainty in momentum

In our case, the electron is assumed to be in the nucleus with radius \(10^{-15} \mathrm{m}\), so the uncertainty in position can be taken as the radius of the nucleus: \[\Delta x = 10^{-15} \mathrm{m}\] Considering the minimum uncertainty, we can calculate the uncertainty in momentum as follows: \[\Delta p = \frac{h}{4\pi\Delta x} = \frac{6.625\times10^{-34} \mathrm{Js}}{4\pi(10^{-15}\mathrm{m})}\]

Calculate the electron velocity using the momentum principle

Using the mass of the electron, denoted by \(m_e=9.1\times10^{-31}\mathrm{kg}\), we can determine the electron's velocity by dividing the momentum by the mass: \[v = \frac{\Delta p}{m_e}\]

Calculate the electron energy

Now that we have the velocity of the electron, we can calculate its energy using the energy-momentum relation: \[E = \frac{1}{2} m_e v^2\]

Convert the energy to Mega Electron-Volt (MeV)

Since the answer choices are given in MeV, we need to convert the energy found in Joules to MeV by dividing the energy by the conversion factor \(1.602\times10^{-13}\mathrm{J/MeV}\): \[E_{MeV}=\frac{E}{1.602\times10^{-13}\mathrm{J/MeV}}\] Plug the values obtained from Steps 2, 3, and 4 into this equation to find the electron's energy in MeV and choose the correct answer from the options provided.