Answer
\( 425 \, \text{nm} \).
Work Step by Step
To find the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal, we can use the equation:
\[ \text{Energy of photon} = \text{Work function} \]
Since we need the energy of the photon to be equal to the work function, we can use the equation:
\[ E = \frac{{hc}}{{\lambda}} \] where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J s),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s), and
- \( \lambda \) is the wavelength of the light.
Rearranging the equation to solve for \( \lambda \), we get:
\[ \lambda = \frac{{hc}}{{E}} \]
Now, plug in the given work function for lithium (\( 279.7 \, \text{kJ/mol} \)) and convert it to joules per electron:
\[ 279.7 \, \text{kJ/mol} \times \frac{{1000 \, \text{J}}}{{1 \, \text{kJ}}} \times \frac{{1 \, \text{mol}}}{{6.022 \times 10^{23} \, \text{atoms}}} \]
\[ = 4.66 \times 10^{-19} \, \text{J/electron} \]
Now, substitute this value for \( E \) in the equation:
\[ \lambda = \frac{{hc}}{{4.66 \times 10^{-19} \, \text{J}}} \]
Calculate:
\[ \lambda = \frac{{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}}{{4.66 \times 10^{-19} \, \text{J}}} \]
\[ \lambda \approx 4.25 \times 10^{-7} \, \text{m} \]
So, the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal is approximately \( 425 \, \text{nm} \).
