1.

Calculate the energy of a photon of blue light, `lambda`=450 nm in air (or vacuum).

Answer» Strategy : The photon has energy E=hf , where `f=c//lambda`
Since `f=c//lambda`, we have
`E=hf="hc"/lambda=((6.63xx10^(-34)J.s)(3.0xx10^8 m//s))/((4.5xx10^(-7)m))=4.4xx10^(-19) `J, or `((4.4xx10^(-19) J))/((1.60xx10^(-19) J//eV))` =2.8 eV
1 eV=`1.60xx10^(-19)` J


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