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(A) The energy of an electronic transition is given by $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.According to the Bohr mode

Vedclass · Accepted Answer

$101.3$

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(A) The energy of an electronic transition is given by $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.According to the Bohr model,the energy of an orbit is $E_n \propto Z^2$. For the same transition (same $n_1$ and $n_2$),the energy difference $\Delta E \propto Z^2$.Therefore,$\frac{1}{\lambda} \propto Z^2$,or $\lambda \propto \frac{1}{Z^2}$.For hydrogen $(H)$,$Z = 1$ and $\lambda_H = 912 \mathring{A}$.For lithium ion $(Li^{2+})$,$Z = 3$.Using the relation $\frac{\lambda_{Li}}{\lambda_H} = \frac{Z_H^2}{Z_{Li}^2}$,we get:$\lambda_{Li} = \lambda_H 	imes \frac{1^2}{3^2} = \frac{912}{9} \mathring{A} = 101.3 \mathring{A}$.

The wavelength corresponding to an electronic transition between two orbits of a hydrogen atom is $912 \mathring{A}$. The wavelength (in $\mathring{A}$) for the same electronic transition in $Li^{2+}$ is

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