Question

The energy of one mole of photons of radiation of wavelength \(300\) \(nm\) is (Given \(h\) = \(6.63 × 10^{–34}\)Js, \(N_A\) = \(6.02 × 10^{23}\) \(mol^{–1}\), \(c = 3 × 10^8\) \(ms^{–1}\))

• A. 235 kJ \(mol^{–1}\)
• B. 325 kJ \(mol^{–1}\)
• C. 399 kJ \(mol^{–1}\)
• D. 435 kJ \(mol^{–1}\)

Answer 1

The Correct Option is C

To find the energy of one mole of photons with a given wavelength, we can use the formulas for energy of a photon and the total energy of a mole of photons.

1. First, calculate the energy of a single photon using the formula: E = \frac{hc}{\lambda}

Where:

• E is the energy of a photon in Joules,
• h is Planck's constant (6.63 \times 10^{-34} \text{ Js}),
• c is the speed of light (3 \times 10^8 \text{ m/s}),
• \lambda is the wavelength in meters (300 \times 10^{-9} \text{ m}).

2. Substitute the given values into the formula: E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}}

Calculate the energy: E = \frac{6.63 \times 3}{300} \times 10^{-19}

E = \frac{19.89}{300} \times 10^{-19} E = 6.63 \times 10^{-19} \text{ J}

3. To find the energy of one mole of photons, multiply the energy of one photon (E) by Avogadro's number (N_A = 6.02 \times 10^{23} \text{ mol}^{-1}):

E_{\text{mole}} = 6.63 \times 10^{-19} \times 6.02 \times 10^{23}

Calculate the total energy: E_{\text{mole}} = 39.9 \times 10^{4} \text{ J/mol} E_{\text{mole}} = 399 \times 10^{3} \text{ J/mol} E_{\text{mole}} = 399 \text{ kJ/mol}

Thus, the energy of one mole of photons of radiation with a wavelength of 300 nm is 399 kJ/mol.

This solution matches with the correct answer option provided, which is 399 kJ/mol.