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
whereh 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
whereh 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
wherec 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)