ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]

Question

contestada

Respuesta :

shallomisaiah19 shallomisaiah19 · Answer 1 · 2020-06-15T02:18:04+02:00

Answer:

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]

Explanation:

Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]

Photon momentum = h/wavelength

[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]

Mass of H2 molecule m = molar mass/Avogadros number

[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]

Since momentum is conserved:

Photon momentum = H2 molecule momentum = mass x velocity of H2

[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]

velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]

Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2

[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]

Translation energy of 1 mole of H2 molecules = KE x Avogadros number

[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]