$(a)\;2 \times 1026 A^{0}\qquad(b)\;2 \times 2052 A^{0}\qquad(c)\;3\; and \;1026 A^{0}\qquad(d)\;3 \times 2052 A ^{0}$

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Answer : (c) $\;3\;and\;1026 A^{0}$

Explanation :

$\varepsilon_{n}^{H}=-\large\frac{13.6}{n^2}\;eV$

$\varepsilon_{1}^{H}=-\large\frac{13.6}{1^2}\;eV=-13.6\;eV \quad \; (n=1)$

$\varepsilon_{2}^{H}=-\large\frac{13.6}{2^2}\;eV=-3.4\;eV \quad \; (n=2)$

$\varepsilon_{3}^{H}=-\large\frac{13.6}{3^2}\;eV=-1.51\;eV \quad \; (n=3)$

Now , $\;\varepsilon_{2}^{H}-\varepsilon _{1}^{H}=-3.4\;eV+13.6eV=10.2 eV$

Now , $\;\varepsilon_{3}^{H}-\varepsilon _{1}^{H}=-1.51\;eV+13.6eV=12.09 eV$

Hence electrons will be raised to the table of principal quantum number 2 and 3 corresponding to radiation energies 10.2eV and 12.09eV respectively . wavelength of radiation emitted for

$\lambda=\large\frac{6.62\times10^{-34}\times3\times10^{8}}{12.09\times1.6\times10^{-19}}=1026 A^{0}\;.$

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