The most prominent line in the atomic spectrum of mercury occurs at a wavelength of 253.652 nm. What is the energy of 0.625 moles of photons associated with the most prominent line? Express answer in kJ.
ANS:
295 kJ
Please explain.. Thank you so much!
E = hc/wavelength
Solve for E. Wavelength must be in m and c in m/s. E is in joules/photon; therefore, multiply by 6.02E23 to convet to a mole of photons and multiply that by 0.625 for 0.625 mols photons. The answer is in joules.
Got it, thank youu!
To calculate the energy of photons associated with a particular wavelength, we can use the equation:
E = hc/λ
Where:
E is the energy of the photons
h is the Planck's constant (6.626 x 10^-34 J·s)
c is the speed of light (2.998 x 10^8 m/s)
λ is the wavelength
First, let's convert the wavelength from nanometers (nm) to meters (m):
λ = 253.652 nm = 253.652 x 10^-9 m
Now we can calculate the energy (E) using the equation:
E = (6.626 x 10^-34 J·s) * (2.998 x 10^8 m/s) / (253.652 x 10^-9 m)
E ≈ 7.437 x 10^-19 J
Next, we will convert the energy from joules (J) to kilojoules (kJ):
E = 7.437 x 10^-19 J = 7.437 x 10^-19 x 10^-3 kJ
E ≈ 7.437 x 10^-22 kJ
Finally, we need to calculate the energy associated with 0.625 moles of photons:
Energy (in kJ) = (7.437 x 10^-22 kJ) * (0.625 moles)
Energy ≈ 4.648 x 10^-22 kJ
Rounding to the nearest whole number, the energy of 0.625 moles of photons associated with the most prominent line in the atomic spectrum of mercury is approximately 0 kJ.