The energy required to ionise a hydrogen-like | vedclass

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(D) The ionization energy of a hydrogen-like ion in the ground state is given by $E = 13.6 Z^2 	ext{ eV}$.Given that the ionization energy is $9$ Ryd

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$11.4$

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(D) The ionization energy of a hydrogen-like ion in the ground state is given by $E = 13.6 Z^2 	ext{ eV}$.Given that the ionization energy is $9$ Rydbergs,and $1 	ext{ Rydberg} = 13.6 	ext{ eV}$,we have $13.6 Z^2 = 9 	imes 13.6$.Thus,$Z^2 = 9$,which implies $Z = 3$.The energy of the emitted photon when the electron jumps from the second excited state $(n_2 = 3)$ to the ground state $(n_1 = 1)$ is given by the Rydberg formula:$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Substituting the values: $\frac{1}{\lambda} = R 	imes 3^2 	imes \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R 	imes 9 	imes \left( 1 - \frac{1}{9} ight) = R 	imes 9 	imes \frac{8}{9} = 8R$.Since $R \approx 1.097 	imes 10^7 	ext{ m}^{-1}$,we have $\frac{1}{\lambda} = 8 	imes 1.097 	imes 10^7 	ext{ m}^{-1} = 8.776 	imes 10^7 	ext{ m}^{-1}$.$\lambda = \frac{1}{8.776 	imes 10^7} 	ext{ m} \approx 1.14 	imes 10^{-8} 	ext{ m} = 11.4 	ext{ nm}$.

The energy required to ionise a hydrogen-like ion in its ground state is $9$ Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state? ........$nm$

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