If the ionization potential of an atom is 122 | vedclass

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Question

(A) The ionization potential $I.P.$ is given by $E_{\infty} - E_1 = 122.4 \, V$. Since $E_n = -\frac{I.P.}{n^2}$,the energy of the ground state is $E_

Vedclass · Accepted Answer

$91.8, 108.8 \, V$

Vedclass · Answer

(A) The ionization potential $I.P.$ is given by $E_{\infty} - E_1 = 122.4 \, V$. Since $E_n = -\frac{I.P.}{n^2}$,the energy of the ground state is $E_1 = -122.4 \, eV$.The energy of the $n^{th}$ state is $E_n = -\frac{122.4}{n^2} \, eV$.The first excited state corresponds to $n = 2$. The excitation potential is $E_2 - E_1 = -\frac{122.4}{4} - (-122.4) = 122.4(1 - \frac{1}{4}) = 122.4 	imes \frac{3}{4} = 91.8 \, V$.The second excited state corresponds to $n = 3$. The excitation potential is $E_3 - E_1 = -\frac{122.4}{9} - (-122.4) = 122.4(1 - \frac{1}{9}) = 122.4 	imes \frac{8}{9} = 108.8 \, V$.

If the ionization potential of an atom is $122.4 \, V$,then the excitation potentials for the first and second excited states are respectively:

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