Question

Use the Bohr model to calculate the radius and the energy of the ${\mathbf{\text{B}}}^{\mathbf{4}\mathbf{+}}$ion in the $\mathbf{\text{n=3}}$ state. How much energy would be required to remove the electrons from 1 mol of ${\mathbf{\text{B}}}^{\mathbf{4}\mathbf{+}}$in this state? What frequency and wavelength of light would be emitted in a transition from the $\mathbf{\text{n=3}}$to $\mathbf{\text{n=2}}$the state of this ion? Express all results in SI units.

Short answer

The radius of ${{\text{B}}^4}^{+}$ is $0.0952\text{\hspace{0.17em}nm}$.

Energy of ${{\text{B}}^4}^{+}$ is $-6.056\times {10}^{-18}\text{\hspace{0.17em}J}$.

Energy required to remove the electrons is $3.65\times {10}^3{\text{\hspace{0.17em}kJmol}}^{-1}$.

The frequency and wavelength of light are $1.143\times {10}^{16}{\text{\hspace{0.17em}s}}^{-1}$ and 26.25 nm, respectively.

Step-by-step solution

Calculation of radius of B4+

An atom is composed of different sub-particles like electrons which revolve around the nucleus in a definite path called shells and each shell has a fixed energy. The electrons gain energy and excite from lower energy level to a higher energy level.

Bohr has introduced an equation to calculate the radius of an ion. This equation is given below:

${\text{r}}_{\text{n}}=\frac{{\text{n}}^2}{\text{Z}}{\text{a}}_0$

n is the number of state, Z is the atomic number, ${\text{a}}_0=\frac{{\text{h}}^2}{4\mathrm{\pi }{\text{me}}^2}=5.29\times {10}^{-11}\text{\hspace{0.17em}m}$, h is the Planck’s constant, m is the mass of electron, e is the charge of electron.

Substituting these values, we get

$\begin{array}{c}{\text{r}}_{\text{n}}=\frac{{\text{3}}^2}{\text{5}}(5.29\times {10}^{-11}\text{m})\\ =9.522\times {10}^{-11}\text{\hspace{0.17em}m=0.0952\hspace{0.17em}nm}\end{array}$

Thus, radius is 0.0952 nm.

Step 2: Calculation of energy of  B4+

Energy of an ion can be calculated as follows:

${\text{E}}_{\text{n}}=\frac{-2{\mathrm{\pi }}^2{\text{me}}^4{\text{Z}}^2}{{\text{n}}^2{\text{h}}^2}{\text{\hspace{0.17em}kJmol}}^{-1}$

This equation can be simplified as given below:

$\begin{array}{c}{\text{E}}_{\text{n}}=\frac{-1312{\text{Z}}^2}{{\text{n}}^2}{\text{\hspace{0.17em}kJmol}}^{-1}\\ =\frac{-2.18\times {10}^{-18}{\text{Z}}^2}{{\text{n}}^2}{\text{\hspace{0.17em}J\hspace{0.17em}atom}}^{-1}\end{array}$

Substituting these values, we get

$\begin{array}{c}{\text{E}}_{\text{3}}=\frac{-2.18\times {10}^{-18}\times {\text{5}}^2}{{\text{3}}^2}\text{\hspace{0.17em}J}\\ =\frac{-2.18\times {10}^{-18}\times 25}{9}\text{\hspace{0.17em}J}\\ =-6.056\times {10}^{-18}\text{\hspace{0.17em}J}\end{array}$

Hence, the energy of ${{\text{B}}^4}^{+}$ ion is $-6.056\times {10}^{-18}\text{\hspace{0.17em}J}$.

Calculation of energy to remove the electrons

When electron is removed, the energy becomes zero. So we need an energy of $+6.056\times {10}^{-18}\text{\hspace{0.17em}J}$.for the removal of electrons. For 1 mole of${\text{B}}^{4+}$ion, the energy change will be the product of Avogadro number and the energy value.

$\begin{array}{c}\mathrm{\Delta }\text{E=}(6.022\times {10}^{23})\times (+6.056\times {10}^{-18})\\ =3.65\times {10}^3{\text{\hspace{0.17em}kJmol}}^{-1}\end{array}$

Thus, the energy needed to remove electrons from 1 mole of ${\text{B}}^{4+}$ ion is $3.65\times {10}^3{\text{\hspace{0.17em}kJmol}}^{-1}$.

Calculation of frequency

Initially, we need to find the energy change of${\text{B}}^{4+}$ion in $\text{n=3}\to \text{2}$ transition. Energy in state $\text{n=2}$ can be calculated as follows:

$\begin{array}{c}{\text{E}}_{\text{2}}=\frac{-2.18\times {10}^{-18}\times {\text{5}}^2}{{\text{2}}^2}\text{\hspace{0.17em}J}\\ =\frac{-2.18\times {10}^{-18}\times 25}{4}\text{\hspace{0.17em}J}\\ =-1.363\times {10}^{-17}\text{\hspace{0.17em}J}\end{array}$

Energy change,

$\begin{array}{c}\mathrm{\Delta }{\text{E=E}}_2-{\text{E}}_3\\ =-1.363\times {10}^{-17}\text{\hspace{0.17em}J-}(-6.056\times {10}^{-18}\text{\hspace{0.17em}J})\\ =-7.574\times {10}^{-18}\text{\hspace{0.17em}J}\end{array}$

We know that

$\text{E=h}\mathrm{\nu }$

Substituting the values, we get the value of frequency.

$\begin{array}{c}\text{-7.574}\times {\text{10}}^{-17}\text{=6.626}\times {\text{10}}^{-34}\times \mathrm{\nu }\\ \mathrm{\nu }=\frac{\text{-7.574}\times {\text{10}}^{-17}}{\text{6.626}\times {\text{10}}^{-34}}\\ =1.143\times {10}^{16}{\text{\hspace{0.17em}s}}^{-1}\end{array}$

Thus, the frequency is $1.143\times {10}^{16}{\text{\hspace{0.17em}s}}^{-1}$ .

Calculation of wavelength

We know that the frequency and wavelength are inter-related as follows:

$\mathrm{\lambda }=\frac{\text{c}}{\mathrm{\nu }}$

$\mathrm{\lambda }$is the wavelength, c is velocity of light, $\nu$ is frequency.

$\begin{array}{c}\mathrm{\lambda }=\frac{(3\times {10}^8{\text{\hspace{0.17em}ms}}^{-1})}{1.143\times {10}^{16}{\text{\hspace{0.17em}s}}^{-1}}\\ =2.625\times {10}^{-8}\text{\hspace{0.17em}m=26.25\hspace{0.17em}nm}\end{array}$

Therefore, the wavelength is 26.25 nm.