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Q.

If the binding energy of the ground state electron in a hydrogen atom is 13.6 eV, then the energy required to remove the electron from the second excited state of Li²⁺ will be; 1 x eV 10^-1. The value of x is _____

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answer is 136.

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Detailed Solution

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Energy required to remove electron $E = \frac{13.62}{n^{2}} eV$
In H-atom $\Rightarrow z = 1$ , ground state n =1

For ${Li}^{2 +} \Rightarrow Z = 3$

2nd excited state , n = 3
$\begin{array}{l} E^{1} = \frac{13.6 \times 3^{2}}{9} = 13.6 eV \\ x \times 10^{- 1} = 13.6 \\ x = 136 \end{array}$

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If the binding energy of the ground state electron in a hydrogen atom is 13.6 eV, then the energy required to remove the electron from the second excited state of Li2+ will be; 1 x eV 10-1. The value of x is _____