1.

The threshold frequency of a metal is 1.11 xx 10^(16) Hz. What is the maximum kinetic energy of the photo electron produced by applying a light of 15Å on the metal?

Answer»

Solution :`v=(C)/(lambda)=(3.0xx10^(8))/(15xx10^(-10)m)=2XX10^(17)Hz`
`KE=hv-hv_(0)`
`KE=[6.625xx10^(-34)Jsxx2xx10^(17)s^(-1)]-[6.625xx10^(-34)Jsxx1.11xx10^(16)s^(-1)]`
`"Kinetic ENERGY "=1.25xx10^(-16)J`


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