$(a)\;385nm\qquad(b)\;585nm\qquad(c)\;185nm\qquad(d)\;485nm$

$\bigtriangleup H(energy)$ = 247KJ/mol (of photon)

$=\large\frac{247\times10^3}{6.02\times10^23}$ j/photon

$=4.103\times10^{-19}$ J/photon

Energy = hv = $\large\frac{hc}{\lambda}$ for a photon

$\therefore \lambda = \large\frac{hc}{\bigtriangleup H}$

$=\large\frac{6.63\times10^{-34} JS\times3\times10^8 m/sec}{4.103\times10^{-19}}$

$=4.85\times10^{-7}m$

=485nm

Hence anwer is (d)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...