Question

The energy required to ionize a hydrogen-like ion in its ground state is $9$ Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?

• A. 8.6
• B. 11.4
• C. 24.2
• D. 35.8

Answer 1

The correct option is B

11.4

Step 1: Given data:

1. Ionization energy = $9Rydbergs$.
2. $1Rydbergs=13.6\mathrm{eV}$

Step 2: Formula used:

Ionization energy $=R{Z}^2$……(1)

According to Bohr's theory, the wavelength can be calculated as,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$
(Where $R$ is Rydberg constant having value $1.0973\times {10}^7{m}^{-1}$,$Z$ is the charge of the nucleus, $n_1,n_2$ are ground and second exciting states respectively.)

Step 3: Calculation of wavelength of an electron,

Ionization energy will be calculated using the formula

$\begin{array}{rcl}9R& =& R{Z}^2\\ Z& =& \sqrt{9}\\ Z& =& 3\end{array}$

Using Bohr's equation, the wavelength can be calculated as-

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$

$n_1=1$ and $n_2=3$

Putting the known values,

$$\begin{gathered} \frac{1}{\lambda }=R\left(3^2\right)(\frac{1}{1^2}-\frac{1}{3^2}) \\ \frac{1}{\lambda }=R\times 9\times \frac{8}{9} \\ \lambda =\frac{1}{8R} \\ \lambda =\frac{1}{8\times 1.0973\times {10}^7} \\ \lambda =11.4\mathrm{nm} \end{gathered}$$

The wavelength of an electron will be $\lambda =11.4nm$

Hence, option B is the correct answer.