Question

Determine that the wavelength of radius 225nm can release an electron from a gold atom or not?

Short answer

Answer

Thus, $8.8346\times {10}^{-19}\mathrm{J}$energy is not sufficient to ionize a gold atom.

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 maximum wavelength

The conversion factor to convert to is as follows:

$1\mathrm{nm}={10}^{-9}\mathrm{m}$

Thus, 225nm is converted to as follows:

$$\begin{gathered} \mathrm{Wavelength}=\left(225\mathrm{nm}\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{mm}}\right) \\ =2.25\times {10}^{-7}\mathrm{m} \end{gathered}$$

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

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

Thus, $890.1\mathrm{kJ}/\mathrm{mol}$is converted to as follows:

$$\begin{gathered} \mathrm{Energy}=\left(890.1\mathrm{kJ}\right)\left(\frac{1000\mathrm{J}}{1\mathrm{kl}}\right) \\ =8.9\times {10}^5\mathrm{J} \end{gathered}$$

Since $8.9\times {10}^3\mathrm{J}$is present for one mole that is for $6.022\times {10}^{23}$atoms so energy per atom is calculated as follows:

$$\begin{gathered} \mathrm{Energy}\mathrm{per}\mathrm{atom}=\left(1\mathrm{atom}\right)\left(\frac{8.5\times {10}^3}{6.002\times {10}^{23}\mathrm{asoms}}\right) \\ =1.477\times {10}^{-18}\mathrm{J}/\mathrm{atom} \end{gathered}$$

Thus to ionize one gold atom the energy required is localid="1656769976429" $1.477\times {10}^{-18}\mathrm{J}$.

The formula to calculate frequency is as follows:

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

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

Value of $\mathrm{\lambda }$is $2.25\times {10}^{-7}\mathrm{m}$.

Substitute the value in the above equation.

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{x}}{\mathrm{i}} \\ =\frac{3\times {10}^2\mathrm{ms}}{225\times {10}^{-1}\mathrm{m}} \\ =1.33\times {10}^{15}{\mathrm{s}}^{-1} \end{gathered}$$

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

$\mathrm{E}=\mathrm{h}-\mathrm{v}$

Value of v is $1.333\times {10}^{15}{\mathrm{s}}^{-1}$.

Value of h is $6.626\times {10}^{-34}\mathrm{Js}$.

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

$$\begin{gathered} \mathrm{E}=\mathrm{h}·\mathrm{v} \\ =\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(1.333\times {10}^{15}{\mathrm{s}}^{-1}\right) \\ =8.8346\times {10}^{-19}\mathrm{J}/\mathrm{photon} \end{gathered}$$

Thus, $8.8346\times {10}^{-19}\mathrm{J}$energy is not sufficient to ionize a gold atom.