1.

For what kinetic energy of a proton, will the associated de-Broglie wavelength be 16.5 nm?

Answer» `lambda=16.5 nm = 16.5 xx 10^(-9)m`.
`m=1.6xx10^(-27)kg`
Using de-Broglie equation, we have
`lambda=(h)/(mv) rArr 16.5xx10^(-9)=(6.626xx10^(-34))/(1.6xx10^(-27)xxv)`
`upsilon = (6.626xx10^(-34))/(16.5xx10^(-9)xx1.6xx10^(-27))=(6.626xx10^(-34))/(26.4xx10^(-36))`
`=0.251xx10^(2)=25.1 m//s`
Hence, kinetic energy `= (1)/(2) mv^(2) = (1)/(2) xx 1.6xx10^(-27)xx(25.1)^(2)`
`=504.008xx10^(-27)=5.04xx10^(-25)J`.


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