How much energy (in kJ) do 4.50 moles of photons contain at a wavelength of 705 nm?

• A. 2.58 x 10² kJ
• B. 7.64 x 10² kJ
• C. 8.45 x 10² kJ
• D. 5.89 x 10² kJ

Answer: (B)

Verified step by step guidance

1. First, understand that the energy of a photon can be calculated using the equation: <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>E</mi><mo>=</mo><mfrac><mi>hc</mi><mi>λ</mi></mfrac></math>, where <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>h</mi></math> is Planck's constant (<math xmlns='http://www.w3.org/1998/Math/MathML'>6.626 × 10^-34 J·s</math>), <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>c</mi></math> is the speed of light (<math xmlns='http://www.w3.org/1998/Math/MathML'>3.00 × 10⁸ m/s</math>), and <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>λ</mi></math> is the wavelength in meters.
2. Convert the wavelength from nanometers to meters. Since 1 nm = 1 × 10^-9 m, the wavelength of 705 nm is <math xmlns='http://www.w3.org/1998/Math/MathML'>705 × 10^-9 m</math>.
3. Substitute the values of <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>h</mi></math>, <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>c</mi></math>, and <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>λ</mi></math> into the energy equation to find the energy of a single photon: <math xmlns='http://www.w3.org/1998/Math/MathML'><mi>E</mi><mo>=</mo><mfrac><mrow><mn>6.626</mn><mo>×</mo><msup><mn>10</mn><mo>-34</mo></msup><mo>×</mo><mn>3.00</mn><mo>×</mo><msup><mn>10</mn><mo>8</mo></msup></mrow><mrow><mn>705</mn><mo>×</mo><msup><mn>10</mn><mo>-9</mo></msup></mrow></mfrac></math>.
4. Calculate the energy for 4.50 moles of photons. Since 1 mole contains <math xmlns='http://www.w3.org/1998/Math/MathML'>6.022 × 10²³</math> photons (Avogadro's number), multiply the energy of one photon by <math xmlns='http://www.w3.org/1998/Math/MathML'>4.50 × 6.022 × 10²³</math>.
5. Convert the total energy from joules to kilojoules by dividing by 1000, as 1 kJ = 1000 J. This will give you the energy in kJ for 4.50 moles of photons.