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What is the ratio of the velocities of `CH_(4)` and `O_(2)` molecules such that they are associated with de broglie waves of equal wavelength? |
Answer» According to de Broglie equation,`lambda=h//m upsilon` For methane `(CH_(4)), " " lambda_(CH_(4))=(h)/(m_(CH_(4))xx upsilon_(CH_(4)))` For oxygen `(O_(2)), " " lambda_(O_(2))=(h)/(m_(O_(2))xxupsilon_(O_(2)))` Since the wavelengths of `CH_(4)` and `O_(2)` are to be equal, `:. " " lambda_(CH_(4))=lambda_(O_(2)) " " "or" (h)/(m_(CH_(4))xx upsilon_(CH_(4)))=(h)/(m_(O_(2))xxupsilon_(O_(2)))` `m_(CH_(4))xx upsilon_(CH_(4))=m_(O_(2))xxupsilon_(O_(2)) " " "or" (upsilon _(CH_(4)))/(upsilon_(O_(2)))=(m_(O_(2)))/(m_(CH_(4)))=(32)/(16)=2`. |
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