Energy of an electron is given by E = -2.178 | vedclass

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(A) The energy difference between two levels is given by $\Delta E = E_2 - E_1 = 2.178 	imes 10^{-18} 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_

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$1.214 	imes 10^{-7} \ m$

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(A) The energy difference between two levels is given by $\Delta E = E_2 - E_1 = 2.178 	imes 10^{-18} 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ J$.For a hydrogen atom,$Z = 1$,$n_1 = 1$,and $n_2 = 2$.$\Delta E = 2.178 	imes 10^{-18} 	imes 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 2.178 	imes 10^{-18} 	imes \left( 1 - 0.25 ight) = 2.178 	imes 10^{-18} 	imes 0.75 = 1.6335 	imes 10^{-18} \ J$.Using the relation $\Delta E = \frac{hc}{\lambda}$,we get $\lambda = \frac{hc}{\Delta E}$.$\lambda = \frac{6.62 	imes 10^{-34} 	imes 3.0 	imes 10^8}{1.6335 	imes 10^{-18}} = \frac{19.86 	imes 10^{-26}}{1.6335 	imes 10^{-18}} \approx 1.216 	imes 10^{-7} \ m$.Rounding to the nearest provided option,the correct answer is $1.214 	imes 10^{-7} \ m$.

Energy of an electron is given by $E = -2.178 \times 10^{-18} \ J \left( \frac{Z^2}{n^2} \right)$. Wavelength of light required to excite an electron in a hydrogen atom from level $n = 1$ to $n = 2$ will be:

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