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The de-Broglie wavelength ofphoton is twice the de-Broglie wavelength of an electron .The speed of the electron is v_(e)=(c )/(100).Then |
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Answer» `(E_(e))/(E_(p))=10^(-4)` (`because lambda_(p)=2lambda_(e)` is given) For ELECTRON `E_(e)=(p_(e)^(2))/(2m_(e))=(1)/(2m_(e))((h^(2))/(lambda_(e)^(2)))` `THEREFORE (E_(e))/(E_(p))=(h^(2))/(2m_(2)lambda_(e)^(2))xx(2lambda_(e))/(hc)=(h)/(m_(e)c lambda_(e))` `therefore (E_(e))/(E_(p))=(h)/(m_(e)c)xx(m_(e)v_(e))/(h)(because lambda_(e)=(h)/(m(e)v_(e)))` `therefore (E_(e))/(E_(p))=(v_(e))/(c)=(1)/(100)(because v_(e)=(c)/(100)` is given) `therefore`(E_(e))/(E_(p))=10^(-2)` `implies` OPTION (B) is correct . Now ,`(p_(e))/(m_(e)c)=(m_(e)v_(e))/(m_(e)c)=(v_(e))/(c)=10^(-2)` `implies Option (C )is also correct. |
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