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

Options

A. `4.44xx10^(-19)J`
B. `2.22xx10^(-19)J`
C. `1.11xx10^(-19)J`
D. `4.80xx10^(-19)J`

Answer

D

Explanation

`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`