Why are de-Broglie waves associated with a moving football not visible ?
The wavelength `lamda` of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of the photon is `(2lamdamc)/(h)` times the kinetic energy of the electron, where m, c and h have their usual meanings.

Explanation

de-Broglie waves associated with a moving football are not visible because in accordance with formula `lamda=(h)/(mv)`, the value of wavelength is extremely small (of the order of `10^(-34)`m).
Energy of a photon of wavelength `lamda, E_(p)=hv=(hc)/(lamda)`
if de-Broglie wavelength of electron be `lamda,` then
`lamda=(h)/(mv) or v=(h)/(mlamda)`
`therefore` K.E. of an electron is `K_(e)=(1)/(2)mv^(2)=(1)/(2)m*((h)/(mlamda))^(2)=(h^(2))/(2mlamda^(2))` . . (ii)
Dividing (i) by (ii), we have
`(E_(p))/(K_(e))=(hc//lamda)/(h^(2)//2mlamda^(2))=(hc)/(lamda)xx(2mlamda^(2))/(h^(2))=(2mlamdac)/(h) or E_(p)=(2mlamdac)/(h)*K_(e)`.