- Identify the relation between energy, frequency, and Planck's constant: \(E = hf\).
- Substitute the provided values: \(h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}\) and \(f = 9.0 \times 10^{10} \text{ Hz}\).
- Perform the multiplication to find the energy: \(E = 5.9634 \times 10^{-23} \text{ J}\).
- State the final energy: \(\boxed{5.9634 \times 10^{-23} \text{ J}}\).
Explanation
- Problem Analysis and Formula Setup
The energy \(E\) of an electromagnetic wave is related to its frequency \(f\) by the Planck-Einstein relation: \[E = hf\] where \(h\) is Planck's constant. The given frequency of the microwave is \(f = 9.0 \times 10^{10} \text{ Hz}\), and Planck's constant is given as \(h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}\).
- Calculation of Energy
To find the energy \(E\), we substitute the given values into the formula: \[E = (6.626 \times 10^{-34} \text{ J} \cdot \text{s}) \times (9.0 \times 10^{10} \text{ Hz})\] First, multiply the coefficients: \(6.626 \times 9.0 = 59.634\). Then, multiply the powers of 10: \(10^{-34} \times 10^{10} = 10^{-34+10} = 10^{-24}\). Combining these gives: \[E = 59.634 \times 10^{-24} \text{ J}\] To express this in standard scientific notation, we adjust the coefficient: \[E = 5.9634 \times 10^{1} \times 10^{-24} \text{ J} = 5.9634 \times 10^{-23} \text{ J}\]
- Final Answer
The energy of the electromagnetic radiation from the microwave is calculated to be \(\boxed{5.9634 \times 10^{-23} \text{ J}}\).
Examples
Understanding the energy of electromagnetic waves is fundamental in fields like telecommunications and medical imaging. For example, by calculating the energy of microwave radiation, engineers can ensure that appliances function efficiently and safely within regulated frequency bands. This calculation is a basic step in analyzing how different types of radiation interact with matter, helping researchers develop everything from modern wireless communication technologies to precise diagnostic medical equipment.