Electrons with a kinetici energy of `6.023xx10^(4)J//mol` are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron fro the metal atom is:

Options

A. `2.3125xx10^(-19)J`
B. `3xx10^(-19)J`
C. `6.02xx10^(-19)J`
D. `6.62xx10^(-34)J`

Answer

A

Explanation

Absorbed energy=Threshold energy+Kinetic energy of photoelectron
`(hc)/(lamda)=E_(0)+KE`
`(6.62xx10^(-34)xx3xx10^(8))/(600xx10^(-9))=E_(0)+(6.023xx10^(4))/(6.023xx10^(23))J//"atom"`
`3.31xx10^(-19)=E_(0)+1xx10^(-19)`
`E_(0)=2.31xx10^(-19)J`