Answer
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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
Complete answer:
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.
Additional information:
Ionization energy: it is also known that ionization potential is the total quotes of energy required to remove electrons from an atom or molecule. Or it is the total energy required for an electron to go into first state from ground state.
Quantum numbers (n):
$\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
Note:
Here as to find ionization potential we have to take the difference of total energy between the first and ground state because it will give the potential of the electron to go in first state from the ground state.
Formula used:
${{\left( T.E \right)}_{F}}=-\dfrac{E}{{{n}^{2}}}$
Where, E = ionisation potential
n = quantum number
Complete answer:
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.
Additional information:
Ionization energy: it is also known that ionization potential is the total quotes of energy required to remove electrons from an atom or molecule. Or it is the total energy required for an electron to go into first state from ground state.
Quantum numbers (n):
$\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
Note:
Here as to find ionization potential we have to take the difference of total energy between the first and ground state because it will give the potential of the electron to go in first state from the ground state.
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