Question

The ionization energy of $L{i}^{2+}$ is equal to:
(A) $9hcR$
(B) $6hcR$
(C) $2hcR$
(D) $hcR$

Answer 1

Hint :Ionization energy is the energy required to remove an electron from the given atom or ion. This is usually expressed in $kJ/mol$ .
Lithium has 3 electrons in total. $L{i}^{2+}$ means that 2 of those 3 electrons are already lost and thus forms a cation. We need to find out the energy needed to remove that last electron from the given cation. We are also asked to express this in terms of $h,c,R$ .
We all know that Energy is given by
 $\Rightarrow E={\displaystyle \frac{hc}{\lambda }}------(1)$
Where $E$ is energy
 $h$ is the Plank’s constant. It has a value of $6.62607004\times {10}^{-34}{m}^{2}kg/s$
 $c$ is the speed of light. It has a value of $3\times {10}^{8}m/s$
 $\lambda$ is the wavelength
We can find the value of $\lambda$ for this question from Rydberg’s formula. It is given by;
 $\Rightarrow {\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}\right)$
 $R$ is known as the Rydberg’s constant. It has a value of $1.097\times {10}^7{m^{-}}^1$
Here $Z$ is the atomic number
 $n_1$ is the initial energy state
 $n_2$ is the final energy state.

Complete Step By Step Answer:
We are asked to find the ionization energy of $L{i}^{2+}$ . This means that we can apply Rydberg’s formula and find ${\displaystyle \frac{1}{\lambda }}$ from it. This can be substituted in equation (1) to find out the energy.
Since we are talking about ionisation from the first energy level of $L{i}^{2+}$ , we have to consider the value of $n_1$ as 1 and that of $n_2$ as $\mathrm{\infty }$ .
We also know that the atomic number of Lithium is 3
Substituting these in the Rydberg’s equation we get:
 $\Rightarrow {\displaystyle \frac{1}{\lambda }}=R(3{)}^2\left({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{{\mathrm{\infty }}_{}^2}}\right)$
Since ${\displaystyle \frac{1}{\mathrm{\infty }}}$ is equal to zero, we can rewrite the equation as:
 $\Rightarrow {\displaystyle \frac{1}{\lambda }}=9R$
Substituting these values into equation (1) we get
 $\Rightarrow E={\displaystyle \frac{hc}{\lambda }}$
 $\Rightarrow E=hc\times 9R$
 $\Rightarrow E=9hcR$
Hence we can say that the correct option for this question is (A).

Note :
We can substitute the values of $h,c,R$ and find out the value of ionization energy of $L{i}^{2+}$ to be $122.4\text{ eV}$
Using Rydberg’s formula we can also be used to find out the wavelength of different spectral lines that arise due to transitions between different energy levels.
They are widely known as Lyman, Balmer, Paschen, Brackett, Pfund.