Question

Photon having wavelength 310 NM is used to break the bond of a two molecule having bond energy 288 kilo joule per mole then the percentage of energy of photon cover to kinetic energy is​

Answer 1

Answer:

Explanation:

Energy of photons = Work function  + Kinetic energy  .......(i)

Work function = 288 KJ

Energy of photon = hc / λ

= 12400 A° / 3100 A°

= 4 ev [ but 1 eV = 96 KJ / mol ]

4 * 96 = 384 KJ / mol

=> By Placing this value in equation (i), we get

384 = 288 + kinetic energy

kinetic energy = 384 - 288 = 96 KJ

Now, kinetic energy % = 96/384 * 100

Kinetic energy % = 1 / 4 * 100 = 25 %

Thus, percentage of energy of photon converted into kinetic energy is 25 %.