Saved Bookmarks
| 1. |
The de Broglie wavelength of an electron moving with a velocity c/2 (c=velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is |
|
Answer» `1:4` `lambda_e = (h)/(pe) = (h)/(m_e (c/2))=(2h)/(m_ec)`.....(i) where `m_e` is the MASS of an electron Kinetic energy of an electron, `K_e =1/2 m_e (C/2)^2 = 1/8 m_e C^2` Kinetic energy of a photon, `K_(ph)=(HC)/(lambdaph)` `because lambda_e = lambda_(ph) `(Given) `THEREFORE K_(ph)= (hc)/(lambda_e)= (hc)/((2h)/(m_eC))=(m_ec^2)/(2)` (Using(i)) `therefore (K_e)/(K_(ph))= 1/8 m_eC^2 xx (2)/(m_ec^2)=1/4` |
|