Question 181 of 949
The energy associated with the photon of a radio transmission at 3 x 105 Hz is
[h = 6.63 x 10-34Js]
- A. 2.00 x 10-28J
- B. 1.30 x 10-28J
- C. 2.00 x 10- 29J
- D. 1.30 x 10-29J
Correct Answer:
A
Explanation
To find the energy associated with a photon of a radio transmission at a frequency of \(3 \times 10^5\) Hz, we can use the formula that relates the energy of a photon to its frequency:
\[
E = h \cdot f
\]
Where:
- \(E\) is the energy of the photon (in joules),
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(f\) is the frequency of the photon (in hertz).
### Step-by-Step Calculation
1. **Identify the values**:
- Frequency, \(f = 3 \times 10^5 \, \text{Hz}\)
- Planck's constant, \(h = 6.63 \times 10^{-34} \, \text{Js}\)
2. **Substitute the values into the formula**:
\[
E = (6.63 \times 10^{-34} \, \text{Js}) \cdot (3 \times 10^5 \, \text{Hz})
\]
3. **Perform the multiplication**:
- First, multiply the coefficients:
\[
6.63 \times 3 = 19.89
\]
- Next, multiply the powers of ten:
\[
10^{-34} \times 10^5 = 10^{-29}
\]
- Combine these results:
\[
E = 19.89 \times 10^{-29} \, \text{J}
\]
4. **Convert to scientific notation**:
- \(19.89\) can be expressed as \(1.989 \times 10^1\), so:
\[
E = 1.989 \times 10^1 \times 10^{-29} = 1.989 \times 10^{-28} \, \text{J}
\]
- Rounding \(1.989\) to two significant figures gives approximately \(2.00\):
\[
E \approx 2.00 \times 10^{-28} \, \text{J}
\]
### Conclusion
The energy associated with the photon of a radio transmission at \(3 \times 10^5\) Hz is approximately \(2.00 \times 10^{-28} \, \text{J}\). Therefore, the correct option is:
**A. 2.00 x 10-28J**
### Explanation of Other Options
- **Option B: 1.30 x 10-28J**: This value is lower than the calculated energy. It likely results from an incorrect multiplication or misunderstanding of the frequency's contribution to energy.
- **Option C: 2.00 x 10-29J**: This value is an order of magnitude lower than the correct answer, indicating a possible error in the calculation of the frequency or Planck's constant.
- **Option D: 1.30 x 10-29J**: Similar to option C, this is also an order of magnitude lower than the correct answer, suggesting a miscalculation.
### Revision Summary
- Use the formula \(E = h \cdot f\) to calculate photon energy.
- Ensure to multiply coefficients and add exponents correctly.
- Convert results to proper scientific notation for clarity.
- Always check calculations to avoid common pitfalls in multiplication and exponent handling.