1.

If energy of photon of 6000 Å is 3.2xx10^(-19) J then energy of photon with 4000 Å will be…….

Answer»

`4.44xx10^(-19)J`
`2.22xx10^(-19)J`
`1.11xx10^(-19)J`
`4.80xx10^(-19)J`

SOLUTION :`E=hf=(hc)/(lambda)`
`THEREFORE E prop (1)/(lambda)[because` hc same]
`therefore (E_(2))/(E_(1))=(lambda_(1))/(lambda_(2))`
`therefore E_(2)=E_(1)xx(lambda_(1))/(lambda_(2))=3.2xx10^(-19)xx(6000)/(4000)`
`therefore E_(2)=4.80xx10^(19)J`


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