If \([Cu(H_2O)_4]^{2+}\) absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [\(Cu(H_2O)_6]^{2+}\) will be _______ \(×10^{–21} J\). [Nearest integer]
(Given : h = \(6.63 × 10^{–34} Js\) and \(c = 3.08×10^8 ms^{–1}\))
\([Cu(H_2O)_4]^{2+}\) is tetrahedral
\([Cu(H_2O)_6]^{2+}\) is octahedral
Because,
\(△_t = \frac 49 \times△_0\)
\(△_t = \frac {6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}\)
\(△_0 = \frac 94 \times \frac {6.63 \times 10^{-34} \times 3.08 \times 10^8}{600 \times 10^{-9}}\)
\(≃ 765 \times 10^{-21}\ J\)
So, the answer is \(765\).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: