Question

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

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, $E_B=-13.6{\displaystyle \frac{Z^2}{n^2}}$ in order to calculate the correct solution.

Formula used:
The formula used in this problem is Rydberg’s formula which is defined as: -
${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}\right)$
where, $R=Rydbergconstant=1.1\times {10}^7{m}^{-1}$, $n_{1}and{n}_2$ are the numbers for low energy level and high energy level respectively.
The expression for Binding Energy is,
$E_B=-13.6{\displaystyle \frac{Z^2}{n^2}}$

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

From the equation $(1)$, we get
${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{5^2}}\right)=R{Z}^2\left({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{25}}\right)=R{Z}^2\left({\displaystyle \frac{21}{100}}\right)$
$Z^2={\displaystyle \frac{1}{R\lambda }}\left({\displaystyle \frac{100}{21}}\right)...(2)$
Substituting the values of $\lambda$ and $R$ in the equation $(2)$ from the question, we get
$Z^2={\displaystyle \frac{1}{\left(1.1\times {10}^7\right)\left(108.5\times {10}^{-9}\right)}}\cdot \left({\displaystyle \frac{100}{21}}\right)$
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: -
$E_B=-13.6{\displaystyle \frac{Z^2}{n^2}}$
For ground state, $n=1$ and substituting $Z=2$ , we get
$E_B=-13.6{\displaystyle \frac{{\left(2\right)}^2}{{\left(1\right)}^2}}$
$\therefore E_B=-13.6\times 4=-54.4eV$
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 $E_B=54.4eV$.

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 $E_B=-13.6{\displaystyle \frac{Z^2}{n^2}}$ 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.