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A photon and a proton have the same de-Broglie wavelength lamda. Prove that the energy of the photon is (2mlamdac//h) times the kinetic energy of the proton. |
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Answer» Solution :We KNOW that, energy ofa photon of WAVELENGTH `lamda," "E_("photon")=hv=(hc)/(lamda)`. . (i) and KINETIC energy of a proton is `K_("proton")=(1)/(2)mv^(2)` If de-Broglie wavelength of proton be `lamda`, then `lamda=(h)/(mv) or v=(h)/(mlamda)` `therefore K_("proton")=(1)/(2)m((h)/(mlamda))^(2)=(h^(2))/(2mlamda^(2))` . . (ii) Dividing (i) by (ii), we have `(E_("photon"))/(K_("proton"))=(hc//lamda)/(h^(2)//2mlamda^(2))=(hc)/(lamda)XX(2mlamda^(2))/(h^(2))=(2mlamdac)/(h)` or `E_("photon")=(2mlamdac)/(h)*K_("proton")`. |
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