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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. For t

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$1.69 	imes 10^{-23} \ eV$

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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$. The energy difference required for excitation is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$. To find the energy per atom (or per mole basis converted to per atom),we divide by the Avogadro number: Energy required = $\frac{10.2 \ eV}{6.022 	imes 10^{23}} \approx 1.69 	imes 10^{-23} \ eV$.

The ionization energy of a hydrogen atom is $-13.6 \ eV$. The energy required to excite an electron from the ground state to the first excited state is ...... (Avogadro number = $6.022 \times 10^{23}$)

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