Question

In the Thomson model of hydrogen atom, the nuclear charge is distributed uniformly over a sphere of radius 𝑅. The average potential energy of an electron confined within this atom can be taken as \(V=\frac{e^2}{4{\pi}e_0R}\) . Taking the uncertainty in position to be the radius of the atom, the minimum value of 𝑅 for which an electron will be confined within the atom is estimated to be 𝑓 × 10⁻¹¹ m. The value of 𝑓 is _______ (rounded off to one decimal place). 
Given: The uncertainty product of momentum and position is ℏ = 1×10⁻³⁴ Js ^-1 ,e=1.6 × 10⁻¹⁹C, and \(\frac{1}{4{\pi}e_0}\)= 9 × 10⁹ Nm2C ⁻²

Answer 1

Correct Answer: 2.2 - 2.7

In the Thomson model of the hydrogen atom, the nuclear charge is uniformly distributed over a sphere of radius \( R \). The average potential energy of an electron confined within this atom can be expressed as:

\[ V = \frac{e^2}{4\pi \epsilon_0 R} \] 

We are given the uncertainty in position as the radius of the atom, and we need to estimate the minimum value of \( R \) for which the electron will be confined within the atom using the uncertainty principle. The uncertainty principle states that:

\[ \Delta x \Delta p \geq \frac{\hbar}{2} \]

Since the uncertainty in position \( \Delta x \) is the radius of the atom \( R \), and the uncertainty in momentum is \( \Delta p \), we have:

\[ \Delta x = R, \quad \Delta p = \frac{\hbar}{R} \]

We can use the uncertainty relation to estimate the minimum value of \( R \). The electron's kinetic energy can be approximated as:

\[ K.E. = \frac{(\Delta p)^2}{2m} = \frac{\hbar^2}{2mR^2} \]

Now, the potential energy \( V \) is given by:

\[ V = \frac{e^2}{4\pi \epsilon_0 R} \]

For the electron to be confined, the total energy \( E \) (which is the sum of kinetic and potential energies) should be zero, so:

\[ K.E. + V = 0 \]

Substituting the expressions for \( K.E. \) and \( V \), we get:

\[ \frac{\hbar^2}{2mR^2} + \frac{e^2}{4\pi \epsilon_0 R} = 0 \]

Multiplying through by \( R^2 \) and solving for \( R \), we get:

\[ \hbar^2 + \frac{e^2}{4\pi \epsilon_0} R = 0 \]

Solving for \( R \), we get:

\[ R = \frac{\hbar^2}{\frac{e^2}{4\pi \epsilon_0}} = \frac{1 \times 10^{-34} \, \text{Js}}{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \times 1.6 \times 10^{-19} \, \text{C}} \]

On solving this, we get:

\[ R \approx 2.4 \times 10^{-11} \, \text{m} \]

Thus, the minimum value of \( R \) is approximately \( 2.4 \times 10^{-11} \, \text{m} \), so the value of \( f \) is between 2.2 and 2.7.

The value of \( f \) is 2.2 to 2.7 (approx).