In Li^{++} ion,an electron in the first Bohr | vedclass

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(A) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state $(n=1)$ is given by the formula: $

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$10.8$

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(A) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state $(n=1)$ is given by the formula: $N = \frac{n(n-1)}{2}$.Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$. Since $n > 0$,the electron is excited to the $n = 4$ level.The energy of the photon absorbed to excite the electron from $n=1$ to $n=4$ in a hydrogen-like ion with atomic number $Z$ is given by: $\Delta E = 13.6 \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) eV$.For $Li^{++}$,$Z = 3$. Thus,$\Delta E = 13.6 	imes 3^2 \left( \frac{1}{1^2} - \frac{1}{4^2} ight) = 13.6 	imes 9 	imes \left( 1 - \frac{1}{16} ight) = 13.6 	imes 9 	imes \frac{15}{16} \approx 114.75 \, eV$.The wavelength $\lambda$ is given by $\lambda = \frac{hc}{\Delta E}$. Using $hc \approx 1240 \, eV \cdot nm$,we get $\lambda = \frac{1240}{114.75} \approx 10.8 \, nm$.

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