If an electron moves from rest from a point at which potential is `50` volt to another point at which potential is `70` volt, then its kinetic energy in the final state will be

Options

A. `3.2 xx 10^(-10)J`
B. `3.2 xx 10^(-18)J`
C. 1 N
D. 1 dyne

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the relationship between potential energy and kinetic energy
The change in kinetic energy (ΔKE) of a charged particle can be related to the change in potential energy (ΔPE) as follows:
\[ \Delta KE = -\Delta PE \]
This means that the change in kinetic energy is equal to the negative change in potential energy.

### Step 2: Calculate the change in potential energy
The potential energy (U) of a charge (Q) in an electric potential (V) is given by the formula:
\[ U = Q \cdot V \]
For an electron, the charge (Q) is approximately \(1.6 \times 10^{-19}\) coulombs. The potential changes from 50 volts to 70 volts, so the potential difference (ΔV) is:
\[ \Delta V = V_f - V_i = 70 \, \text{V} - 50 \, \text{V} = 20 \, \text{V} \]

### Step 3: Calculate the change in potential energy
Using the formula for potential energy:
\[ \Delta PE = Q \cdot \Delta V \]
Substituting the values:
\[ \Delta PE = (1.6 \times 10^{-19} \, \text{C}) \cdot (20 \, \text{V}) = 3.2 \times 10^{-18} \, \text{J} \]

### Step 4: Relate change in potential energy to change in kinetic energy
Since the change in kinetic energy is equal to the change in potential energy:
\[ \Delta KE = \Delta PE \]
Thus, we have:
\[ \Delta KE = 3.2 \times 10^{-18} \, \text{J} \]

### Step 5: Conclusion
The kinetic energy of the electron in the final state will be:
\[ KE = 3.2 \times 10^{-18} \, \text{J} \]

### Final Answer
The kinetic energy in the final state will be \(3.2 \times 10^{-18} \, \text{J}\).

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