An electron in an atom jumps in such a way that its kinetic energy changes from x to `x/4`. The change in potential energy will be:

Options

A. `+3/2x`
B. `-3/8x`
C. `+3/4x`
D. `-3/4x`

Answer

To solve the problem of the change in potential energy when an electron's kinetic energy changes from \( x \) to \( \frac{x}{4} \), we can follow these steps:

### Step 1: Calculate the Change in Kinetic Energy
The initial kinetic energy (KE_initial) is \( x \) and the final kinetic energy (KE_final) is \( \frac{x}{4} \). The change in kinetic energy (ΔKE) can be calculated as:

\[ \Delta KE = KE_{final} - KE_{initial} = \frac{x}{4} - x \]

### Step 2: Simplify the Change in Kinetic Energy
To simplify the expression:

\[ \Delta KE = \frac{x}{4} - \frac{4x}{4} = \frac{x - 4x}{4} = \frac{-3x}{4} \]

### Step 3: Relate Kinetic Energy to Total Energy
According to the virial theorem, in a bound system, the total energy (E) is related to kinetic energy (KE) and potential energy (PE) as follows:

\[ E = KE + PE \]

Since we know that the total energy is also related to the kinetic energy, we can express the total energy in terms of the change in kinetic energy.

### Step 4: Calculate Total Energy
If we assume that the potential energy is zero initially (for simplicity), the total energy when the kinetic energy is \( x \) would be:

\[ E_{initial} = KE_{initial} + PE_{initial} = x + 0 = x \]

After the jump, the total energy with the new kinetic energy is:

\[ E_{final} = KE_{final} + PE_{final} = \frac{x}{4} + PE_{final} \]

### Step 5: Set Up the Equation for Total Energy
Using the virial theorem, we know that:

\[ PE_{final} = 2 \times E_{final} \]

Substituting \( E_{final} \):

\[ PE_{final} = 2 \times \left(\frac{x}{4} + PE_{final}\right) \]

### Step 6: Solve for Potential Energy
Rearranging the equation gives:

\[ PE_{final} = \frac{x}{2} + 2 \times PE_{final} \]

This leads to:

\[ PE_{final} - 2 \times PE_{final} = \frac{x}{2} \]

\[ -PE_{final} = \frac{x}{2} \]

Thus:

\[ PE_{final} = -\frac{x}{2} \]

### Step 7: Calculate Change in Potential Energy
Now, we can find the change in potential energy (ΔPE):

\[ \Delta PE = PE_{final} - PE_{initial} = -\frac{x}{2} - 0 = -\frac{x}{2} \]

### Conclusion
The change in potential energy when the electron jumps is:

\[ \Delta PE = -\frac{x}{2} \]