- #1
stunner5000pt
- 1,443
- 4
Determine from classical mechanics (using a head-on collision with recoil at 180 degrees) what fraction of an electron’s kinetic energy can be transferred to a mercury atom in an elastic collision. Derive an approximate value of the fraction.
well ok let the mass of the electron be m_e
initial velocity of electron v_1
fina lvelocity v_2
mercury atom mass m_Hg
mercury atom fina lvelocity v_Hg
\frac{1}{2} m_{e} v_{1}^2 = \frac{1}{2} m_{Hg} v_{Hg}^2 + \frac{1}{2} m_{e} v_{2}^2
the fraction of kinetic energy is Kf/Ki right
\frac{K_{f}}{K_{i}} = \frac{v_{2}^2}{v_{1}^2}
do i use momentum concepts to find v2 in terms of v1
still it doesn't give me a numerical value... if that's what the question is asking...
well ok let the mass of the electron be m_e
initial velocity of electron v_1
fina lvelocity v_2
mercury atom mass m_Hg
mercury atom fina lvelocity v_Hg
\frac{1}{2} m_{e} v_{1}^2 = \frac{1}{2} m_{Hg} v_{Hg}^2 + \frac{1}{2} m_{e} v_{2}^2
the fraction of kinetic energy is Kf/Ki right
\frac{K_{f}}{K_{i}} = \frac{v_{2}^2}{v_{1}^2}
do i use momentum concepts to find v2 in terms of v1
still it doesn't give me a numerical value... if that's what the question is asking...
