Question

By using photons of specific wavelengths, chemists can dissociate gaseous HI to produce H atoms with certain speeds. When HI dissociates, the H atoms move away rapidly, whereas the heavier I atoms move more slowly. (a) If a photon of 240. nm is used, what is the excess energy (in J) over that needed for dissociation? (b) If this excess energy is carried away by the H atom as kinetic energy, what is its speed (in m/s)? (Bond energy of H―I is 295 kJ/mol.) Enter your answers in scientific notation.

Answer 1

(a) Energy of photon = 240 nm = 240 *10^-9 m

Now, E = hc/

Where , h = planck's constant = 6.626 *10^-34 J.s

c = velocity of light = 3*10⁸ m/s

= 240 *10^-9 m

E = (6.626 *10^-34 J.s *  3*10⁸ m/s)/  240 *10^-9 m = 8.28 *10^-19 J

Now, bond enegy of HI = 295 kJ/mol = 295 *10³ J/mol

Thus, E = bond energy / Avogadros number =  295 *10³ J/mol/ 6.022*10²³ = 4.898*10^-19 J

Hence, excess energy over dissociation =8.28 *10^-19 J -  4.898*10^-19 J = 3.382*10^-19 J

(b) Mass of Hydrogen = mass of proton + mass of electron

= 9.1 *10^-31 kg + 1.67 *10^-27kg = 1.67 *10^-27 kg

Kinetic energy = mv²/2 =1.67 *10^-27 kg *V²/2

3.382*10^-19 J = 1.67 *10^-27 kg *V²/2

V = 2.01*10⁴ m/s

Hence, speed = 2.01*10⁴ m/s