Question #279865

Show that energy transferred by energetic electron to helium atom by elastic collision is less than 0.1%.

1
Expert's answer
2021-12-20T10:29:07-0500

Take the frame of reference at a fixed helium nucleus\text{Take the frame of reference at a fixed helium nucleus}

m1helium atom m_1 - \text{helium atom }

m2electronm_2 - \text{electron}

v1=0v_1 = 0

EK=m2v222EK = \frac{m_2v_2^2}{2}

m2v2=m1u1+m2u2m_2\vec{v_2}=m_1\vec{u_1} +m_2\vec{u_2}

m1u1=m2v2m2u2m_1\vec{u_1} = m_2\vec{v_2}-m_2\vec{u_2}

max(u1)=2v2m2m1max (\vec u_{1})= \frac{2\vec v_2m_2}{m_1}


ΔEKm1=m1u122\Delta EK_{m_1}= \frac{m_1u_1^2}{2}

max(ΔEKm1)=2m22v22m1max(\Delta EK_{m_1})= \frac{2m_2^2v_2^2}{m_1}

maxpercentage=max(ΔEKm1)EK100=4m2m1100\text{maxpercentage}= \frac{max(\Delta EK_{m_1})}{EK}*100= \frac{4m_2}{m_1}*100

The helium nucleus - is made up of two protons and two neutrons.\text{The helium nucleus - is made up of two protons and two neutrons.}

The mass of a proton is 1836 heavier than the mass of an electron\text{The mass of a proton is 1836 heavier than the mass of an electron}

maxpercentage==4m241836m2100=0.054%\text{maxpercentage}= = \frac{4m_2}{4*1836*m_2}*100= 0.054 \%

maxpercentage<0.1%\text{maxpercentage}<0.1 \%


Proven\text{Proven}


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