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# The first excitation potential of a given atom is 10.2 V then the ionisation potential is:\begin{align} & A.\text{ }10.2V \\ & B.\text{ }13.6V \\ & C.\text{ }30.6V \\ & D.\text{ }20.4V \\ \end{align}

Last updated date: 17th Jun 2024
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Hint: First we will find quantum number (n) and by using quantum number we will find total energy on first state by using the formula $\dfrac{E}{{{n}^{2}}}$ after that we will take difference between first and second state so that we can find ionisation potential (E)
Formula used:
${{\left( T.E \right)}_{F}}=-\dfrac{E}{{{n}^{2}}}$
Where, E = ionisation potential
n = quantum number

Given that
First excitation potential is 10.2V
Now the quantum number on first state is n = 2
Now the total energy in ground state is
${{\left( T.E \right)}_{g}}=-E$
Total energy in first state is
${{\left( T.E \right)}_{F}}=-\dfrac{E}{{{n}^{2}}}$
Now,
$\therefore$The potential of given atom = ${{\left( T \right)}_{F}}-{{\left( T.E \right)}_{g}}$
\begin{align} & \Rightarrow 10.2=-\dfrac{E}{{{n}^{2}}}-\left( -E \right) \\ & \Rightarrow 10.2=-\dfrac{E}{{{n}^{2}}}+E \\ & \Rightarrow 10.2=E-\dfrac{E}{4} \\ & \Rightarrow 10.2=\dfrac{4E-E}{4} \\ & \Rightarrow 40.8=3E \\ & \therefore E=13.6V \\ \end{align}

Hence option (b) 13.6V is correct.

$\to$ There are four quantum numbers which are used to describe the location of electrons in an atom.
$\to$ In this section we will talk about the principal quantum number (n) it is used to describe the shell or electron shell of an atom whose value is ranging from 1 to the outer shell that is different from the different material.
$\to$The principal quantum number (n) describes the size of the orbital. For example if Orbital’s quantum number n = 2 then it is larger than those for which n = 1