Loading...
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.
← Previous Next →
Jump to: 181 182 183 184 185 186 187 188 189 190