The mass of an electron is `9.11xx10^(-31)` kg and the velocity of light is `3.00xx10^(8)ms^(-1)`. Calculate the energy of the electron using Einstein's mass energy relation `E=mc^(2)`, correct ot the significant figures?

Explanation

To solve the problem of calculating the energy of an electron using Einstein's mass-energy relation \( E = mc^2 \), we will follow these steps:

### Step 1: Identify the given values
- Mass of the electron \( m = 9.11 \times 10^{-31} \) kg
- Speed of light \( c = 3.00 \times 10^{8} \) m/s

### Step 2: Write down the formula
The formula we will use is:
\[ E = mc^2 \]

### Step 3: Substitute the values into the formula
Substituting the given values into the equation:
\[ E = (9.11 \times 10^{-31} \, \text{kg}) \times (3.00 \times 10^{8} \, \text{m/s})^2 \]

### Step 4: Calculate \( c^2 \)
First, calculate \( c^2 \):
\[ c^2 = (3.00 \times 10^{8})^2 = 9.00 \times 10^{16} \, \text{m}^2/\text{s}^2 \]

### Step 5: Substitute \( c^2 \) back into the energy equation
Now substitute \( c^2 \) back into the equation:
\[ E = 9.11 \times 10^{-31} \, \text{kg} \times 9.00 \times 10^{16} \, \text{m}^2/\text{s}^2 \]

### Step 6: Perform the multiplication
Now, multiply the numbers:
\[ E = 9.11 \times 9.00 \times 10^{-31} \times 10^{16} \]
Calculating \( 9.11 \times 9.00 \):
\[ 9.11 \times 9.00 = 81.99 \]
Thus,
\[ E = 81.99 \times 10^{-15} \, \text{J} \]

### Step 7: Convert to scientific notation
Convert \( 81.99 \times 10^{-15} \) to proper scientific notation:
\[ E = 8.199 \times 10^{-14} \, \text{J} \]

### Step 8: Round to significant figures
Since both \( 9.11 \) and \( 3.00 \) have three significant figures, we will round \( 8.199 \) to three significant figures:
\[ E \approx 8.20 \times 10^{-14} \, \text{J} \]

### Final Answer
The energy of the electron is:
\[ E = 8.20 \times 10^{-14} \, \text{J} \]

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