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When one mole of electrons de-excites all producing photons with a wavelength of 650 nm how much energy in kilojoules is released

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Final answer:

When one mole of electrons de-excites, it releases approximately 3.04 x 10^-22 kilojoules of energy.


Step-by-step explanation:

When one mole of electrons de-excites, it means that the electrons transition from higher energy levels to lower energy levels, releasing energy in the form of photons. The energy released can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.

Converting the wavelength 650 nm to meters (650 nm = 6.50 x 10^-7 m) and substituting the values into the equation, we get:

E = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s) / (6.50 x 10^-7 m) = 3.04 x 10^-19 J

To convert the energy from joules to kilojoules, divide by 1000:

E = 3.04 x 10^-19 J / 1000 = 3.04 x 10^-22 kJ

Therefore, when one mole of electrons de-excites, it releases approximately 3.04 x 10^-22 kilojoules of energy.


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