An element undergoes a reaction as shown : X + 2 e - → X 2 - , energy released = 30 .87 eV / atom . If the energy released is used to dissociate 4 gms of H 2 molecules, equally into H + and H * , where H * is excited state of H atoms where the electron travels in orbit whose circumference equal to four times its de Broglie's wavelength. Determine the least moles of X that would be required: Given: I.E. of H = 13 .6 e

Step-by-step solution

Total energy requirement per molecule of H 2 : Binding energy =4 .526 eV I .E = 13 .60 eV According to de Broglie's explanation of Bohr's orbit- 2πr n = nλ As per question, 2πr = 4λ Thus, electron is excited to the 4 th Bohr orbit Excitation energy = E 4 − E 1 = − 0 .85 − − 13 .60 = 12 .75 eV Thus, total eenergy requirement per molecule = 13 .60 + 12 .75 + 4 .526 ≅ 30 .87 eV Thus, 1 molecule of H 2 will be dissociated by 1 atom of X Therefore, 4 gH 2 ≡ 2 moles of H 2 molecules will be dissociated by 2 moles of X .