An electron in an atom jumps in such a way that its kinetic energy changes from `x` to `x/9`. The change in its potential energy (magnitude) will be-

Options

A. `x/9`
B. `(16x)/9`
C. `9x`
D. `(9x)/16`

Answer

To solve the problem, we need to determine the change in potential energy when the kinetic energy of an electron changes from \( x \) to \( \frac{x}{9} \).

### Step-by-Step Solution:

1. **Calculate the Change in Kinetic Energy:**
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{x}{9} - x \]
To simplify this, we can write:
\[ \Delta KE = \frac{x}{9} - \frac{9x}{9} = \frac{x - 9x}{9} = \frac{-8x}{9} \]
Thus, the change in kinetic energy is:
\[ \Delta KE = -\frac{8x}{9} \]

2. **Determine the Magnitude of Change in Kinetic Energy:**
Since we are interested in the magnitude:
\[ |\Delta KE| = \frac{8x}{9} \]

3. **Relate Kinetic Energy to Potential Energy:**
According to Bohr's model, the kinetic energy (\( KE \)) and potential energy (\( PE \)) of an electron in an atom are related by:
\[ KE = -\frac{1}{2} PE \]
Therefore, the change in potential energy can be expressed in terms of the change in kinetic energy.

4. **Calculate the Change in Potential Energy:**
The change in potential energy (\( \Delta PE \)) is given by:
\[ \Delta PE = -2 \times \Delta KE \]
Substituting the value of \( |\Delta KE| \):
\[ \Delta PE = -2 \times \left(-\frac{8x}{9}\right) = \frac{16x}{9} \]

5. **Final Answer:**
Thus, the magnitude of the change in potential energy is:
\[ |\Delta PE| = \frac{16x}{9} \]

### Summary:
The change in potential energy (magnitude) when the kinetic energy changes from \( x \) to \( \frac{x}{9} \) is \( \frac{16x}{9} \).

---