An electron and a photon have same wavelength #lambda#, what is the ratio of their kinetic energies ??
2 Answers
Check for error below. See the other solution.
Explanation:
We know that de-Broglie wavelength
#lambda=h/p#
where#h# is Planck's Constant.
Kinetic energy is given as
#KE=p^2/(2m)#
In terms of de-Broglie wavelength
#KE=(h/lambda)^2/(2m)#
#=>KE=h^2/(2mlambda^2)#
Ratio of kinetic energies of electron and proton
#(KE_e)/(KE_p)=(h^2/(2mlambda^2))_e/(h^2/(2mlambda^2))_p#
When both have same wavelength above reduces to
#(KE_e)/(KE_p)=m_p/m_e#
I misread
Explanation:
We know that de-Broglie wavelength
#lambda=h/p#
where#h# is Planck's Constant.
Kinetic is given as
#KE=p^2/(2m)#
In terms of de-Broglie wavelength
#KE=(h/lambda)^2/(2m)#
#=>KE=h^2/(2mlambda^2)#
We know that photon is a mass less particle. Its Kinetic Energy is same as its Energy which is given by the expression
#E=(hc)/lambda#
where#c# is velocity of light
Ratio of kinetic energies of electron and photon
#(KE_e)/(KE_p)=(h^2/(2m_elambda_e^2))/((hc)/lambda_p)#
When both have same wavelength
#(KE_e)/(KE_p)=h/(2m_elambdac)#
