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A beam of hydrogen molecules travels in the z direction with a kinematic energy of 1eV....
A beam of hydrogen molecules travels in the z direction with a kinematic energy of 1eV. The molecules are in an excited state, from which they decay and dissociate in to two hydrogen atoms. When one of the dissociated atoms has its final velocity perpendicular to the z direction its kinetic energy is always 0.8eV. What is the energy released in the dissociate reaction.
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We know that momentum of the moving atom can be expressed as:
`p = sqrt(2mE)`
`p_0` = momentum of `H_2`
`p_0 =sqrt(2mE) `
`p_0 =sqrt(2*2*938E^6 *1) `
`p_0 = 6.125E^4 (eV)/c `
`p_1` = momentum of the first H atom
`p_1 =sqrt(2mE) `
`p_1 =sqrt(2*1*938E^6 *0.8) `
`p_1 = 3.874E^4 (eV)/c `
The momentum of the second hydrogen atom can no be written as:
`p_2 = sqrt(p_0 ^2 + p_1 ^2) `
Substituting the values of p_1 and p_0
`p_2 = sqrt((6.125E^4)^2 + (3.874E^4)^2) `
`p_2 = 7.248E4(eV)/c `
The same particle would have a kinetic energy equal to:
`E_2 = ((7.248E^4 )^2)/(2*938^6) `
`E_2 = 2.80 eV `
Finally, the energy released in the dissociative reaction is:
0.8 + 2.8 - 1 = 2.6 eV -> final answer
Posted by jerichorayel on June 26, 2013 at 5:38 AM (Answer #1)
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