Question

Theratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom is:

• A. 4
• B. \( \frac{1}{4} \)
• C. \(\frac{1}{2}\)
• D. 1

Answer 1

The Correct Option is C

The problem asks for the ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom.

Concept Used:

In the Bohr model of the hydrogen atom, the total energy of an electron in the \(n^{th}\) orbit is the sum of its kinetic energy (K.E.) and potential energy (P.E.). The relationships between these energies are derived from the balance between the electrostatic force and the centripetal force.

The centripetal force required for circular motion is provided by the electrostatic attraction between the nucleus and the electron:

\[ \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \]

From this, we can derive the expressions for K.E. and P.E.:

Kinetic Energy (K.E.):

\[ K.E. = \frac{1}{2}mv^2 = \frac{1}{8\pi\epsilon_0} \frac{e^2}{r} \]

Potential Energy (P.E.):

\[ P.E. = -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \]

By comparing the expressions, we can establish a direct relationship between K.E. and P.E.:

\[ P.E. = -2 \times (K.E.) \]

This relationship is valid for any orbit \(n\).

Step-by-Step Solution:

Step 1: Write down the standard expressions for the kinetic energy (K.E.) and potential energy (P.E.) of an electron in any orbit \(n\) of a hydrogen atom.

The kinetic energy is given by:

\[ K.E. = \frac{1}{2}mv^2 \]

From the force balance equation, \(mv^2 = \frac{e^2}{4\pi\epsilon_0 r}\), so:

\[ K.E. = \frac{e^2}{8\pi\epsilon_0 r} \]

The electrostatic potential energy is given by:

\[ P.E. = -\frac{e^2}{4\pi\epsilon_0 r} \]

Step 2: Find the magnitude of the potential energy, \(|P.E.|\).

\[ |P.E.| = \left| -\frac{e^2}{4\pi\epsilon_0 r} \right| = \frac{e^2}{4\pi\epsilon_0 r} \]

Step 3: Form the ratio of the magnitude of the kinetic energy to the magnitude of the potential energy.

The kinetic energy is always positive, so its magnitude is just K.E. The required ratio is:

\[ \frac{K.E.}{|P.E.|} = \frac{\frac{e^2}{8\pi\epsilon_0 r}}{\frac{e^2}{4\pi\epsilon_0 r}} \]

Step 4: Simplify the expression for the ratio.

The common terms \( \frac{e^2}{4\pi\epsilon_0 r} \) cancel out:

\[ \frac{K.E.}{|P.E.|} = \frac{1/2}{1} = \frac{1}{2} \]

This shows that the ratio of the kinetic energy to the magnitude of the potential energy is constant for any energy level of the hydrogen atom. The fact that the electron is in the 5th excited state (which corresponds to the principal quantum number \(n=6\)) does not change this fundamental ratio.

Final Computation & Result:

The ratio is calculated as:

\[ \frac{\text{Magnitude of Kinetic Energy}}{\text{Magnitude of Potential Energy}} = \frac{K.E.}{|P.E.|} = \frac{1}{2} \]

The ratio is 1:2.

Answer 2

For an electron in a hydrogen atom (Bohr model):

\[ |PE| = 2 \times KE \]

So,

\[ \frac{KE}{|PE|} = \frac{1}{2} \]