The de-Broglie wavelength of an electron moving with a velocity `1.5xx10^(8) ms^(-1)` is equal to that of a photon. Caculate the ratio of the kinetic energy of the electron to that of photon.

Answer

`1//4`

Explanation

Here `v=1.5xx10^(8)ms^(-1)`
de-broglie wavelength of electrons,
`lambda=m/(mv) or m=h/(vlambda)`
K.E of electron, `K_(e)=-1/2mv^(2)=1/2(h/(v lambda))v^(2)=(hv)/(2lambda)`
K.E. of photon, `K_(p)=(hc)/lambda`
`:. (K_(e))/(K_(p))=(hv//2lambda)/(hc//lambda)=v/(2c)=(1.5xx10^(8))/(2xx(3xx10^(8)))=1/4`