The ionization energy of a hydrogen atom is 1 | vedclass

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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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$1.63 	imes 10^{-18} \ J$

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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 required for excitation is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$. To convert this energy into Joules,we multiply by the conversion factor $1.602 	imes 10^{-19} \ J/eV$: $\Delta E = 10.2 	imes 1.602 	imes 10^{-19} \ J \approx 1.634 	imes 10^{-18} \ J$.

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to excite the electron in a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is (Note: The question asks for energy per atom in Joules. Given $1 \ eV = 1.602 \times 10^{-19} \ J$)

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