The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the `5^(" th ")`excited state of a hydrogen atom is :

Options

A. 1
B. `1/2`
C. `1/4`
D. 4

Answer

To find the ratio of the magnitude of the kinetic energy (K.E) to the potential energy (P.E) of an electron in the 5th excited state of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Energy Formulas
For a hydrogen atom, the potential energy (U) and kinetic energy (K) of an electron in a given orbit can be expressed as:
- Potential Energy (U) = - (k * Z^2 * e^2) / r
- Kinetic Energy (K) = (k * Z * e^2) / (2r)

Where:
- k is Coulomb's constant,
- Z is the atomic number (for hydrogen, Z = 1),
- e is the charge of the electron,
- r is the radius of the orbit.

### Step 2: Determine the Radius for the 5th Excited State
The radius of the nth orbit in a hydrogen atom is given by:
\[ r_n = n^2 \cdot a_0 \]
Where \( a_0 \) is the Bohr radius (approximately \( 5.29 \times 10^{-11} \) m). For the 5th excited state, n = 5:
\[ r_5 = 5^2 \cdot a_0 = 25 \cdot a_0 \]

### Step 3: Calculate Kinetic Energy and Potential Energy
Substituting r into the formulas for K and U:
- Kinetic Energy (K):
\[ K = \frac{k \cdot Z \cdot e^2}{2 \cdot r} = \frac{k \cdot e^2}{2 \cdot (25 \cdot a_0)} = \frac{k \cdot e^2}{50 \cdot a_0} \]

- Potential Energy (U):
\[ U = -\frac{k \cdot Z^2 \cdot e^2}{r} = -\frac{k \cdot e^2}{25 \cdot a_0} \]

### Step 4: Find the Ratio of Kinetic Energy to Potential Energy
Now, we can find the ratio of the magnitudes of K.E to P.E:
\[ \text{Ratio} = \frac{|K|}{|U|} = \frac{\frac{k \cdot e^2}{50 \cdot a_0}}{\frac{k \cdot e^2}{25 \cdot a_0}} \]

### Step 5: Simplify the Ratio
This simplifies to:
\[ \text{Ratio} = \frac{1/50}{1/25} = \frac{25}{50} = \frac{1}{2} \]

### Final Answer
Thus, 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 is:
\[ \frac{1}{2} \]