Based on the equation Delta E = - 2.0 times 1 | vedclass

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(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{2^2} -

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$1.325 	imes 10^{-7} \, m$

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(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{2^2} - \frac{1}{1^2} ight)$$\Delta E = - 2.0 	imes 10^{-18} 	imes \left( \frac{1}{4} - 1 ight) = - 2.0 	imes 10^{-18} 	imes \left( - \frac{3}{4} ight) = 1.5 	imes 10^{-18} \, J$Using the relation $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is:$\lambda = \frac{hc}{\Delta E} = \frac{6.625 	imes 10^{-34} 	imes 3 	imes 10^8}{1.5 	imes 10^{-18}}$$\lambda = 1.325 	imes 10^{-7} \, m$

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