If the binding energy of the electron in a hydrogen atom is $ 13.6eV $ then the energy required to remove the electron from the first excited state of $ L{i^{ + + }} $ is
(A) $ 122.4eV $
(B) $ 30.6eV $
(C) $ 13.6eV $
(D) $ 3.4eV $
Answer
591k+ views
Hint
To solve this question, we have to use the formula for the total energy of an electron in the nth orbit. For removing the electron from an atom, we have to make its total energy zero.
Formula Used: The formula used in solving this question is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}} $
Here $ {E_n} $ is the energy in electron-volts for an electron in the nth state of an atom having atomic number equal to $ z $.
Complete step by step answer
We know that the energy of an electron in the nth state of an atom is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}} $ ………………….(1)
As the atom given in the question is Lithium, which has an atomic number of $ 3 $, so we have $ z = 3 $.
Also, the ground state corresponds to $ n = 1 $. So the first excited state for an atom will correspond to the $ n = 2 $ state. So we have $ n = 2 $. Substituting these values in (1), we get
$\Rightarrow {E_n} = - 13.6\dfrac{{{3^2}}}{{{2^2}}} $
$\Rightarrow {E_n} = - 30.6eV $
Therefore the energy of the electron in the first excited state of $ L{i^{ + + }} $ is equal to $ - 30.6eV $.
For removing the electron, we have to make it free from the effect of the nucleus. That is, we need to make its total energy zero. For this, let the energy supplied to the electron be equal to $ E $. So we have
$\Rightarrow {E_n} + E = 0 $
$\Rightarrow - 30.6 + E = 0 $
So we get the energy supplied as
$\Rightarrow E = 30.6eV $
Thus the energy which is required to remove the electron from the first excited state of $ L{i^{ + + }} $ is equal to $ 30.6eV $.
Hence, the correct answer is option B.
Note
If we remember the formula for the total energy of an electron in the nth state, then we do not require the value of the binding energy of the electron in the hydrogen atom, which is given in the question. But if we do not remember it, then we can use this value, as it corresponds to the energy in the ground state of hydrogen $ n = 1 $, which has $ z = 1 $.
To solve this question, we have to use the formula for the total energy of an electron in the nth orbit. For removing the electron from an atom, we have to make its total energy zero.
Formula Used: The formula used in solving this question is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}} $
Here $ {E_n} $ is the energy in electron-volts for an electron in the nth state of an atom having atomic number equal to $ z $.
Complete step by step answer
We know that the energy of an electron in the nth state of an atom is given by
$\Rightarrow {E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}} $ ………………….(1)
As the atom given in the question is Lithium, which has an atomic number of $ 3 $, so we have $ z = 3 $.
Also, the ground state corresponds to $ n = 1 $. So the first excited state for an atom will correspond to the $ n = 2 $ state. So we have $ n = 2 $. Substituting these values in (1), we get
$\Rightarrow {E_n} = - 13.6\dfrac{{{3^2}}}{{{2^2}}} $
$\Rightarrow {E_n} = - 30.6eV $
Therefore the energy of the electron in the first excited state of $ L{i^{ + + }} $ is equal to $ - 30.6eV $.
For removing the electron, we have to make it free from the effect of the nucleus. That is, we need to make its total energy zero. For this, let the energy supplied to the electron be equal to $ E $. So we have
$\Rightarrow {E_n} + E = 0 $
$\Rightarrow - 30.6 + E = 0 $
So we get the energy supplied as
$\Rightarrow E = 30.6eV $
Thus the energy which is required to remove the electron from the first excited state of $ L{i^{ + + }} $ is equal to $ 30.6eV $.
Hence, the correct answer is option B.
Note
If we remember the formula for the total energy of an electron in the nth state, then we do not require the value of the binding energy of the electron in the hydrogen atom, which is given in the question. But if we do not remember it, then we can use this value, as it corresponds to the energy in the ground state of hydrogen $ n = 1 $, which has $ z = 1 $.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

