Answer:
The longest wavelength of radiation that passesses the necessary energy for breaking the Cl- Cl bond (in Cl2) is approximately 494.2 nm, which corresponds to the visible spectrum.
Explanation:
In order to answer this question we need to recall that the energy of a photon is given by:
E = hc/lambda, where
E = energy
h = Planck's constant
c = speed of light in vacuum
lambda = associated photon wavelength
In order to perform the calculations, first we need to change the units of 242kJ/mol to J. For doing this, we to divide by Avogadro's number and multiply by a 1000:
242kJ/mol = (242kJ/mol)*(1mol/6.022x10^23 particles)*(1000J/1kJ)= 4.0186x10^-19 J
Now, we simply solve for lambda and substitute the appropriate values in the energy equation:
lambda = hc/E = (6.626x10^-34 J s)*(3x10^8 m/s)/(4.0186x10^-19 J) = (1.986x10^-25 J m)/(4.0186x10^-19 J) = 4.942x10^-7 m = 494.2x10^-9 m = 494.2 nm
Therefore, the wavelength for a photon to break the Cl-Cl bond in a Cl2 molecule should be 494.2 nm at most, which corresponds to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).
