If change in energy (Delta E) = 3 times 10^{- | vedclass

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(C) The relationship between energy and wavelength is given by the formula: $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength: $\lambda = \fr

Vedclass · Accepted Answer

$6.64 	imes 10^{-8} \ \mathring{A}$

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(C) The relationship between energy and wavelength is given by the formula: $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength: $\lambda = \frac{hc}{\Delta E}$.Substituting the given values: $\lambda = \frac{6.64 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{3 	imes 10^{-8} \ J}$.$\lambda = 6.64 	imes 10^{-18} \ m$.Since $1 \ m = 10^{10} \ \mathring{A}$,we have $\lambda = 6.64 	imes 10^{-18} 	imes 10^{10} \ \mathring{A} = 6.64 	imes 10^{-8} \ \mathring{A}$.

If change in energy $(\Delta E) = 3 \times 10^{-8} \ J$,$h = 6.64 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$,then the wavelength of the light is:

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