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 one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: