Answer
No
Work Step by Step
To determine if light with a wavelength of \( 225 \mathrm{~nm} \) is capable of ionizing a gold atom, we can use the relationship between energy and wavelength given by the equation:
\[ E = \frac{{hc}}{{\lambda}} \]
where:
- \( E \) is the energy of the light,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J s),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength of the light.
Substituting the given values:
\[ E = \frac{{(6.626 \times 10^{-34} \mathrm{~J\cdot s})(3.00 \times 10^8 \mathrm{~m/s})}}{{225 \times 10^{-9} \mathrm{~m}}} \]
\[ E \approx 8.81 \times 10^{-19} \mathrm{~J} \]
Now, we convert this energy into kilojoules per mole to compare it with the ionization energy of gold:
\[ 8.81 \times 10^{-19} \mathrm{~J} \times \frac{{1 \mathrm{~kJ}}}{{1000 \mathrm{~J}}} \times \frac{{6.022 \times 10^{23} \mathrm{~atoms}}}{{1 \mathrm{~mol}}} \]
\[ \approx 530.5 \mathrm{~kJ/mol} \]
The energy of the light is much lower than the ionization energy of gold (\( 890.1 \mathrm{~kJ/mol} \)). Therefore, light with a wavelength of \( 225 \mathrm{~nm} \) is not capable of ionizing a gold atom in the gas phase.
