Question

Light of wavelengths shorter than 275 nm can be used to photodissociate the hydrogen molecule into hydrogen atoms in the gas phase. A 10.0 mL glass cylinder contains H2(g) at 70.0 mtorr and 25 °C. What minimum amount of light energy must be absorbed by the hydrogen in the tube to dissociate 48.0% of the molecules?

energy absorbed:

J

Answer 1

Light have wavelength,id = 275 pm = 276 X10 Pm Energy of photon is given by, Er the where h= plank is constant = 6+63x10 347 c= speed of light = 3x108 m/s a. Energy of 1 photon = 6.63 X 10 34 Js x 3 x 188 m/s 975 X 109m le = 7. 23 x 1019 5 Moles of Hydrogen can be calculated as :- PV = nRT n = pv -5 760 Preis wre given = 70 m tour = 70X163 - 9.28105 T-25°C = 298k V=10ml = 0 DIOL Number of moles, n = 9. 2x10 satmx 0. 0p 0.08206 t-atm k'-mol'x 298k n = 3.734 10 8 moles Number of mokcules = number of moles & Avodaguo's number = 3.73 81058 x 60d2 X1o23 = 22.46 X1075 - 2.25 x 100 molecules.

Number of photons can be calculated as a Number of photons = 48 X 2. 25 x 100 molecules = 1008 x 10's photons Energy of I photon = 7. 33 x 10 195 Energy of (1.08 X10% photons) = (1.23 41019 J x 1.08 xido photons] I E = 7.8088103 J = 70818103