Ionisation energy of $He^{+}$ is 19.6$ imes 10^{- 18}J\ atom^{- 1}$. The energy of the first stationary state ($n = 1$) of $Li^{2 +}$ is

Question

Ionisation energy of $He^{+}$ is 19.6$	imes 10^{- 18}J\ atom^{- 1}$. The energy of the first stationary state ($n = 1$) of $Li^{2 +}$ is

Dr. Abubacker · Accepted Answer

(b)
IE=$- E_{1}$
$E_{1}$ for $He^{+} = - 19.6 	imes 10^{- 18}J\ atom^{- 1}$
$\frac{\left( E_{1} ight)_{He^{+}}}{\left( E_{1} ight)_{Li^{2} + \ }} = \frac{\left( Z_{He^{+}} ight)^{2}}{\left( Z_{Li^{2 +}} ight)^{2}}$
$\frac{- 19.6 	imes 10^{- 18}}{\left( E_{1} ight)_{Li^{2 +}}} = \frac{4}{9}$
$or\ \ E_{1}\left( Li^{2 +} ight) = \frac{- 19.6 	imes 9 	imes 10^{- 18}}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - 4.41 	imes 10^{- 17}J\ atom^{- 1}$

Q. Ionisation energy of $He^{+}$ is 19.6$\times 10^{- 18}J\ atom^{- 1}$. The energy of the first stationary state ($n = 1$) of $Li^{2 +}$ is

Explanation

Reviewed by Dr. Abubacker

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