Q. Energy of an electron is given by $\hspace15mm E=-2.178 \times 10^{-18} \, J$ Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be $ \, \, \, (6.62 \times 10^{-34} \, Js \, and \, \, c=3.0 \times 10^8 \, ms^{-1})$

Solution:

Given, in the question E=-2.178 $\times 10^{-18} \, J \big[\frac{Z^2}{n^2}\big]$ For hydrogen Z = 1 So, $\hspace15mm E_1 =-2.178 \times 10^{-18} \, J \big[\frac{1}{1^2}\big]$ $\hspace20mm E_2=-2.178 \times 10^{-18} \, J \big[\frac{1}{2^2}\big]$ Now, $E_1 - E_2$ i.e. $\hspace15mm \Delta E =2.178 \times 10^{-18} \big(\frac{1}{1^2}- \frac{1}{2^2} \big)=\frac{hc}{\lambda}$ $2.178 \times 10^{-18}\big(\frac{1}{1^2}- \frac{1}{2^2} \big)=\frac{6.62 \times 10^{-34} \times 3.0 \times 10^8}{\lambda}$ $\therefore \hspace35mm \lambda \approx 1.21 \times 10^{-7} \, m$

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