Question

$O_2$ undergoes photochemical dissociation into one normal and one excited atoms. The excited atom has 1.967 eV more energy than normal. The dissociation of $O_2$ into two normal atoms requires 498 KJ mole-1. Then the maximum wavelength required for photochemical dissociation of $O_2$ is

• A. 174nm
• B. 7420A⁰
• C. 742A⁰
• D. 1740A⁰

Answer 1

The correct option is

C

1740A⁰

We know as per the Mole concept, one mole of atoms =$6.022\times {10}^{23}$ atoms

So, energy of one atom can be found out by dividing one by this number.

$E_{peratom}=\frac{498\times {10}^3}{6.023\times {10}^{23}}J$

= $8.26\times {10}^{-19}$

$E_{forexitation}=1.967\times 1.6\times {10}^{-19}$

= $3.1392\times {10}^{-19}$

total $E$ = $8.26\times {10}^{-19}$

$\frac{3.13\times {10}^{-19}}{11.39\times {10}^{-19}J}$

$E=\frac{hc}{\lambda }$

$\lambda$ = $\frac{hc}{E}=\frac{6.625\times {10}^{-34}3\times {10}^8}{11.39\times {10}^{-19}}$

= $1.74\times {10}^{-7}\times {10}^{-10}$

= $1740\AA$

For the calculation part, first, we multiply the upper term and then divide it by the lower term, that is how we are getting this answer, which is correct.