`O_(2)` undergoes photochemical dissociation into one normal oxygen atom and one excited oxygen atom. Excited oxygen atom is `1.967 eV` more energetic than normal . The dissociation of `O_(2)` into two normal atoms of oxygen required `498 kJ mol^(-1)`, what is the maximum wavelength effective for photochemical dissociation of `O_(2)`?

Options

A. `1.01 nm`
B. `1.64 nm`
C. `1.74 nm`
D. `2.74 nm`

Answer

C

Explanation

`O_(2) rarr O +O^(**)`, Dissociation energy `=468 kJ mol^(-1)=(498xx10^(3))/(6.02xx10^(23)) J mol^(-1)=8.27xx10^(-19) J molecules^(-1)`
Extra energy of the excited atom `=1.967 eV=1.967xx1.6xx10^(-19) J=3.15xx10^(-19) J atom^(-1)`
`lambda=(hc)/(E)=(6.626xx10^(-24)xx3xx10^(8))/(11.42xx10^(-19))=174xx10^(-9) m`
Thus, `lambda=174 nm`