Question

The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number $Z$ equals to $108.5\;nm$ . The binding energy of the electron in the ground state of these ions is ${E_B}$ ​. Then
A. $Z = 2$
B. ${E_B} = 54.4\,eV$
C. $Z = 3$
D. ${E_B} = 122.4\,eV$

Answer 1

Hint:In this problem, to determine the binding energy of the electron in the ground state of hydrogen-like ions; we will apply Rydberg's formula to find the value of $Z$ and then substitute this value in the expression of Binding Energy, ${E_B} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}$ in order to calculate the correct solution.

Formula used:
The formula used in this problem is Rydberg’s formula which is defined as: -
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
where, $R = \;Rydberg{\text{ }}constant = 1.1 \times {10^7}\,{m^{ - 1}}$, ${n_1}{\text{ }}and{\text{ }}{n_2}$ are the numbers for low energy level and high energy level respectively.
The expression for Binding Energy is,
${E_B} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}$

Complete step by step solution:
We know that the expression for Rydberg’s formula can be stated as: -
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1) \\ $
Now it is given that the third line of the Balmer series spectrum of a hydrogen-like ion equals $108.5\;nm$. Therefore, for the Balmer series spectrum for the given ion ${n_1} = 2$ and ${n_2} = 5$. Also,
$\lambda = 108.5\;\,nm = 108.5 \times {10^{ - 9}}\,m$..........$\left( {\because 1\,nm = {{10}^{ - 9}}\,m} \right)$

From the equation $(1)$, we get
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{5^2}}}} \right)\, = R{Z^2}\left( {\dfrac{1}{4} - \dfrac{1}{{25}}} \right) = R{Z^2}\left( {\dfrac{{21}}{{100}}} \right)$
${Z^2} = \dfrac{1}{{R\lambda }}\left( {\dfrac{{100}}{{21}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(2) \\ $
Substituting the values of $\lambda $ and $R$ in the equation $(2)$ from the question, we get
${Z^2} = \dfrac{1}{{\left( {1.1 \times {{10}^7}} \right)\left( {108.5 \times {{10}^{ - 9}}\,} \right)}} \cdot \left( {\dfrac{{100}}{{21}}} \right) \\ $
On simplifying it, we get
${Z^2} = 3.9896 \approx 4$
$ \Rightarrow Z = 2$
which means a hydrogen-like ion used in this spectrum is nothing but the $H{e^ + }$ ion.

Now, the expression for Binding Energy can be stated as: -
${E_B} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}} \\ $
For ground state, $n = 1$ and substituting $Z = 2$ , we get
${E_B} = - 13.6\dfrac{{{{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^2}}} \\ $
$\therefore {E_B} = - 13.6 \times 4 = - 54.4\,eV$
Here, Negative binding energy indicates a spectrum’s degree of stability. i.e., a spectrum is more stable if the binding energy is negative. Thus, the binding energy of the electron in the ground state of hydrogen-like ions of atomic number $Z = 2$ will be ${E_B} = 54.4\,eV$.

Hence, the correct options are A and B.

Note: In this problem, first Rydberg’s formula is used to calculate the atomic number of ions and then apply ${E_B} = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}$ to determine the binding energy of the electron in the ground state. Also, the key points like ${n_1} = 2$ (for Balmer series of lines) and $n = 1$ (for ground state) must be kept in mind while doing the calculation part.