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

Concept Used:

Within the Bohr model of the hydrogen atom, an electron's total energy in the \(n^{th}\) orbit is the sum of its kinetic energy (K.E.) and potential energy (P.E.). These energy relationships are derived from the equilibrium between electrostatic and centripetal forces.

The electrostatic attraction between the nucleus and the electron provides the centripetal force necessary for circular motion:

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

From this equation, the expressions for K.E. and P.E. can be derived:

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.):

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

Comparing these expressions reveals a direct relationship between K.E. and P.E.:

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

This relationship holds true for any electron orbit \(n\).

Step-by-Step Solution:

Step 1: State the standard formulas 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 defined as:

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

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

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

The electrostatic potential energy is given by:

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

Step 2: Calculate 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: Compute the ratio of the magnitude of the kinetic energy to the magnitude of the potential energy.

Since kinetic energy is always positive, its magnitude is 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 ratio expression.

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 indicates that the ratio of kinetic energy to the magnitude of potential energy is constant for all energy levels in a hydrogen atom. The specific state, the 5th excited state (corresponding to \(n=6\)), does not alter this fundamental ratio.

Final Computation & Result:

The ratio is computed 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.