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An electron and proton are possessing the same amount of kinetic energy. Which of the two has greater wavelength? |
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Answer» Solution :`K=(1)/(2)mv^(2)=(p^(2))/(2M), LAMBDA=(h)/(p), p=(h)/(lambda), therefore K=(h^(2))/(2m lambda^(2))` KE of electron, `K_(e )=(h^(2))/(2m_(e )lambda_(e )^(2))` KE of proton, `K_(p)=(h^(2))/(2m_(p)lambda_(p)^(2))` Given, `K_( e)=K_(p)` `(h^(2))/(2m_(e )lambda_(e )^(2))=(h^(2))/(2m_(p)lambda_(p)^(2))` `(lambda_(e ))/(lambda_(p))=sqrt((m_(p))/(m_(e )))` `m_(p) GT m_(e )`, Hence, `lambda_(e ) gt lambda_(p)`. |
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