Question

Calculate the Energy of X rays and radio waves in kJ/mol .

Short answer

Answer

Energy of ratio waves$=11.96\times {10}^{-9}\mathrm{kJ}/\mathrm{mol}$

Step-by-step solution

Concept used

The formula to calculate frequency is as follows:

$\mathbf{\lambda }\mathbf{·}\mathbf{y}\mathbf{=}\mathbf{r}$

Where

- $\mathbf{\lambda }$denotes wavelength.

- v denotes frequency.

- c denotes the speed of light.

As per Planck's hypothesis, the energy associated with a photon is given as follows:

$\mathbf{E}\mathbf{=}\mathbf{h}\mathbf{·}\mathbf{v}$

Where,

- v denotes frequency.

- h denotes Planck's constant.

Avogadro's number is$\mathbf{6}\mathbf{.}\mathbf{0222}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$atoms mol.

Calculate the energy of radio waves

The wavelength of x rays is $1\times {10}^{-10}\mathrm{m}$.

The wavelength of radio waves is$1\times {10}^4\mathrm{m}$.

The total energy emitted by C sample is$1.98\times {10}^3\mathrm{J}$.

The formula to calculate frequency is as follows:

$\mathrm{v}=\frac{\mathrm{e}}{2}$

For X-rays

Value of $\mathrm{\lambda }$is $1\times {10}^{-10}\mathrm{m}$.

Value of c is $3\times {10}^8\mathrm{m}/\mathrm{s}$.

Substitute the value in the above equation

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{c}}{2} \\ =\frac{3\times {10}^5\mathrm{ms}}{1\times {10}^{-10}\mathrm{m}} \\ =3\times {10}^{18}{\mathrm{s}}^{-1} \end{gathered}$$

For radio waves

Value of $\mathrm{\lambda }$is $1\times {10}^4\mathrm{m}$.

Value of c is $3\times {10}^5\mathrm{m}/\mathrm{s}$.

Substitute the value in the above equation.

As per Planck's hypothesis, the energy associated with a photon is given as follows:

$\mathrm{E}=\mathrm{h}·\mathrm{v}$

For X-rays

The value of v is $3\times {10}^{18}{\mathrm{s}}^{-1}$.

The value of h is $6.626\times {10}^{-34}\mathrm{J}$\%

Substitute the value in the above equation to calculate the energy for a single photon.

$$\begin{gathered} \mathrm{E}=\mathrm{h}·\mathrm{v} \\ =\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(3\times {10}^{18}{\mathrm{s}}^{-1}\right) \\ =1.987\times {10}^{-15}\mathrm{J} \end{gathered}$$

Energy for 1 mol X -rays is calculated as follows:

localid="1656774050527" $$\begin{gathered} \mathrm{E}=\left(6.022\times {10}^{23}\mathrm{atoms}/\mathrm{mol}\right)\left(1.987\times {10}^{-15}\mathrm{J}/\mathrm{atom}\right) \\ =11.96\times {10}^8\mathrm{J}/\mathrm{mol} \end{gathered}$$

Energy for 1 mol radio waves is calculated as follows:

$$\begin{gathered} \mathrm{E}=\left(6.022\times {10}^{23}\mathrm{atoms}/\mathrm{mol}\right)\left(1.987\times {10}^{-2}\mathrm{J}/\mathrm{atom}\right) \\ =11.96\times {10}^6\mathrm{J}/\mathrm{mol} \end{gathered}$$

The conversion factor to convert J to kJ is as follows:

$\mathrm{lJ}={10}^{-3}\mathrm{kJ}$

Thus, $11.96\times {10}^8\mathrm{J}/\mathrm{mol}$is converted to m as follows:

$$\begin{gathered} \mathrm{Energy}\mathrm{of}\mathrm{X}-\mathrm{rays}=\left(11.96\times {10}^{\mathrm{s}}\mathrm{J}/\mathrm{mol}\right)\left(\frac{{10}^{-3}\mathrm{k}}{1\mathrm{J}}\right) \\ =11.96\times {10}^5\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Similarly $11.96\times {10}^{-6}\mathrm{J}/\mathrm{mol}$is converted to m as follows:

$$\begin{gathered} \mathrm{Energy}\mathrm{of}\mathrm{radio}-\mathrm{waves}=\left(11.96\times {10}^6\mathrm{J}/\mathrm{mol}\right)\left(\frac{{10}^{-3}\mathrm{k}}{1\mathrm{J}}\right) \\ =11.96\times {10}^{-9}\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Since carbon-carbon single disruption requires 347kJ/mol and therefore X rays can cause disruption of such bonds but radio waves cannot cause disruption of such bonds.