γ
-ray photons with a wavelength of  2.56×10−5nm
. Express the energy numerically in kilojoules per mole.

To calculate the energy of the gamma-ray photons, we'll use the following relationship derived from Planck's equation:

\[
E = \frac{hc}{\lambda}
\]

Where:
- \( E \) is the energy of a single photon.
- \( h \) is Planck's constant: \( 6.626 \times 10^{-34} \, \text{J·s} \).
- \( c \) is the speed of light: \( 3.0 \times 10^8 \, \text{m/s} \).
- \( \lambda \) is the wavelength of the gamma-ray photons, which is given as \( 2.56 \times 10^{-5} \, \text{nm} = 2.56 \times 10^{-14} \, \text{m} \) (since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \)).

### Steps:
1. Calculate the energy of a single photon in joules.
2. Convert the energy of one photon to kilojoules.
3. Multiply by Avogadro's number (\( 6.022 \times 10^{23} \)) to get the energy per mole in kilojoules.

Let's perform the calculations.The energy of the gamma-ray photons is approximately \( 4.68 \times 10^9 \, \text{kJ/mol} \).

Would you like a detailed explanation of any part of the calculation or have further questions?

Here are 5 related questions for further practice:
1. How does the energy of a photon change with wavelength?
2. What is the relationship between energy and frequency for a photon?
3. How would the energy change if the wavelength were halved?
4. How do gamma rays compare in energy to other types of electromagnetic radiation?
5. How is Planck's constant used in quantum mechanics?

**Tip:** Shorter wavelengths correspond to higher energy photons due to the inverse relationship in Planck's equation.

γ-ray photons with a wavelength of 2.56×10−5 nm. Express the energy numerically in kilojoules per mole.

Calculate the energy of γ-ray photons with a wavelength of 2.56×10−5 nm using Planck's equation. The step-by-step solution includes determining photon energy in joules, converting it to kilojoules, and calculating the energy per mole using Avogadro's number. Suitable for Grades 11-12 and college-level physics students.

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