Consider one He^{+} ion is in excited state ( | vedclass

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(D) For a single $He^{+}$ ion in the $n = 5$ state,the number of emission lines is $n - 1 = 4$. Thus,option $A$ is incorrect.The ionization energy for

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The electronic separation from the centre of nucleus is more than $6 \, \mathring{A}$

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(D) For a single $He^{+}$ ion in the $n = 5$ state,the number of emission lines is $n - 1 = 4$. Thus,option $A$ is incorrect.The ionization energy for $n = 5$ is given by $I.E. = \frac{13.6 	imes Z^2}{n^2} = \frac{13.6 	imes 2^2}{5^2} = \frac{13.6 	imes 4}{25} = 2.176 \, eV$. Thus,option $B$ is incorrect.The longest wavelength corresponds to the smallest energy transition,which is $n = 5 	o n = 4$. The formula is $\frac{1}{\lambda} = R 	imes Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight) = R 	imes 4 	imes \left(\frac{1}{4^2} - \frac{1}{5^2}ight) = 4R 	imes \left(\frac{25-16}{400}ight) = 4R 	imes \frac{9}{400} = \frac{9R}{100}$. Therefore,$\lambda = \frac{100}{9R} \approx \frac{11.11}{R}$. Since $11.11/R > 10/R$,option $C$ is incorrect.The radius of the orbit is given by $r_n = 0.529 	imes \frac{n^2}{Z} \, \mathring{A}$. For $n = 5$ and $Z = 2$,$r_5 = 0.529 	imes \frac{25}{2} = 0.529 	imes 12.5 = 6.6125 \, \mathring{A}$. Since $6.6125 \, \mathring{A} > 6 \, \mathring{A}$,option $D$ is correct.

Consider one $He^{+}$ ion is in excited state $(n = 5)$. Which of the following observations hold true as per the Bohr's model?

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