If a photon and an electron have the same de-Broglie wavelength of `0.5 Å`, then find the ratio of kinetic energy of photon to that of electron (Mass of electron `=9.1 xx 10^(-31) kg`)

Explanation

Wavelength of photon= Wavelength of an electron `=lamda= 0.5 Å`
`lamda= 0.5 xx 10^(-10)m`
We know that de-Broglie wavelength of an electron `=lamda= (h)/(sqrt(2mE))`
where E= Kinetic energy of electron
`rArr lamda^(2)= (h^(2))/(2mE)`
`E= (h^(2))/(2m lamda^(2))`
Also kinetic energy of the photon is given by `E.= (hc)/(lamda)`
Now `(E.)/(E )= (hc)/(lamda) xx (2m lamda^(2))/(h^(2))= (2m c lamda)/(h)`
`h= 6.62 xx 10^(-34) Js, c= 3 xx 10^(8) ms^(-1), m= 9.1 xx 10^(-31) kg`
`rArr (E.)/(E )= (2xx 9.1 xx 10^(-31) xx 3 xx 10^(8) xx 0.5 xx 10^(-10))/(6.62 xx 10^(-34))`
`(E.)/(E )= (41.4)/(1)= 41.4`