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In Ques, 126, ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest electron emitted is : |
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Answer» 100 `E=(hc)/(lambda)` `:. lambda=(hc)/(E)=(6.63xx10^(-34)xx3xx10^(3))/(5.0xx1.6xx10^(-19))` `=2486xx10^(-10)m` If v is velocity of fastest photoelectron, then Using `E=w+(1)/(2) mv^(2)` We find `v= sqrt((3.2xx10^(-19)xx2)/(m))` Then momentum of photoelectron `mv=m sqrt((3.2xx10^(-19)xx2)/(m))` `= sqrt(3.2xx10^(-19)xx2m)=7.63xx10^(-25)kg m//s` de-Brogli wavelength of the tastest electrons `lambda_(D)=(h)/(mv)=(6.63xx10^(-34))/(7.63xx10^(-25))=8.689xx10^(-10)m` `:. "Required ratio"(lambda)/(lambda_(D))=(2468xx10^(-10))/(8.689xx10^(-10))=286` |
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