1.

An electron and a photon have the same de Broglie wavelength. Which one of these has higher kinetic energy?

Answer» For electron. Let `lambda` be the de -Broglie wavelength of electron.
K.E. of electron, `E_1=1/2mv^2 or mv^2=2E_1`
or `mv=sqrt(2E_1m)`
As `lambda=h/(mv) :. lambda=h/(sqrt(2E_1m))`
or `E_1=(h^2)/(2 lambda^2m)`
For photon of wavelength `lambda`, Energy is given by,
`E_2=(hc)/lambda`
`:. (E_2)/(E_1)=(hc)/lambdaxx(2lambda^2m)/(h^2)=(2clambdam)/h`
`=(2xx3xx10^8xx10^(-10)xx8xx10^(-31))/(6.6xx10^(-34))`
`=(2xx3xx9xx10)/6.6=90/1.1gt1` ltbegt Hence, `E_2gtE_1`
It means kinetic energy of photon is greater than that of electron.


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