What is the energy content per photon (J) for light of frequency `4.2xx10^(14)` Hz?

Options

A. `2.8xx10^(-21)`
B. `2.5xx10^(-19)`
C. `2.8xx10^(-19)`
D. `2.5xx10^(-18)`

Answer

To calculate the energy content per photon for light of frequency \(4.2 \times 10^{14}\) Hz, we can use the formula derived from Planck's equation:

### Step-by-Step Solution:

1. **Identify the formula**: The energy \(E\) of a photon can be calculated using the formula:
\[ E = h \nu \]
where:
- \(E\) is the energy of the photon (in joules),
- \(h\) is Planck's constant (\(6.626 \times 10^{-34}\) Joule seconds),
- \(\nu\) is the frequency of the light (in Hz).

2. **Substitute the values**: We know the frequency \(\nu = 4.2 \times 10^{14}\) Hz. Now, substituting the values into the formula:
\[ E = (6.626 \times 10^{-34} \, \text{J s}) \times (4.2 \times 10^{14} \, \text{Hz}) \]

3. **Perform the multiplication**: Calculate the energy:
\[ E = 6.626 \times 4.2 \times 10^{-34} \times 10^{14} \]
First, calculate \(6.626 \times 4.2\):
\[ 6.626 \times 4.2 = 27.8292 \]
Now, combine the powers of ten:
\[ E = 27.8292 \times 10^{-20} \, \text{J} \]

4. **Convert to scientific notation**: To express \(27.8292 \times 10^{-20}\) in proper scientific notation:
\[ E = 2.78292 \times 10^{-19} \, \text{J} \]
Rounding this to two significant figures gives:
\[ E \approx 2.8 \times 10^{-19} \, \text{J} \]

5. **Final answer**: Therefore, the energy content per photon for light of frequency \(4.2 \times 10^{14}\) Hz is:
\[ E \approx 2.8 \times 10^{-19} \, \text{J} \]