One quantum is absorbed per gaseous molecule of Bromine 2 for converting into Bromine atoms. If light absorbed has wavelength 5000 angstrom, then the bond energy of Bromine 2 is about KJ/mol. Find KJ/mol.
Hi Dhiraj, use the expression E = h c / L
E = energy per quantum meant for one molecule
h - Planck's constant = 6.626 x 10^-34 J s
c - velo. of light - 3 x 10^8 m/s
L - lambda - wavelength - 5000 x 10^-10 m
So, for 1 g mol ie 6.023 x 10^23 (N) molecules energy needed is N * E
Now 6.023 x 10^23 x 6.626 x 10^-34 x 3 x 10^8 /5000 x 10^-10 J./ g mol
= 2,394 x 10^11 J / g mol
Or 2.394 x 10^8 kJ / g mol
E = energy per quantum meant for one molecule
h - Planck's constant = 6.626 x 10^-34 J s
c - velo. of light - 3 x 10^8 m/s
L - lambda - wavelength - 5000 x 10^-10 m
So, for 1 g mol ie 6.023 x 10^23 (N) molecules energy needed is N * E
Now 6.023 x 10^23 x 6.626 x 10^-34 x 3 x 10^8 /5000 x 10^-10 J./ g mol
= 2,394 x 10^11 J / g mol
Or 2.394 x 10^8 kJ / g mol
