Question

$\left[Ti\right(H_{2}O{)}_6{]}^{3+}$ absorbs light of wavelength 498 nm during a d–d transition. The octahedral splitting energy for the above complex is $\times {10}^{-19}J$. (Round off to the nearest integer). $h=6.626\times {10}^{-34}\text{J s}$; c = 3 \times 10^8~\text{ m s}^{-1}\).

Answer 1

${\Delta }_0=\frac{hc}{I_{abs}}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{498\times {10}^{-9}}$
$=\frac{6.626\times 3}{498}\times {10}^{-17}=3.99\times {10}^{-19}$
Nearest integer $=4\times {10}^{-19}J$