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it requires a minimum of 216.4 kj/mol to remove an electron from a potassium metal surface. if uv light at 129 nm strikes this surface, what is the maximum speed (in m/s) of the ejected electrons?
m/s
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Explanation:

Step1: Calculate the energy of the incident photon

First, use the formula $E = \frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c = 3\times 10^{8}\ m/s$ (speed of light), and $\lambda=129\ nm = 129\times 10^{-9}\ m$.
$E=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{129\times 10^{-9}}\ J$
$E = 1.54\times 10^{-18}\ J$ per photon.
For 1 mole of photons, multiply by Avogadro's number $N_A=6.022\times 10^{23}\ mol^{-1}$.
$E_{mol}=1.54\times 10^{-18}\times6.022\times 10^{23}\ J/mol = 927388\ J/mol=927.388\ kJ/mol$.

Step2: Calculate the kinetic energy of the ejected electron

The work - function $\phi = 216.4\ kJ/mol=216400\ J/mol$.
The kinetic energy of the ejected electron per mole $K_{mol}=E_{mol}-\phi$.
$K_{mol}=927388 - 216400=710988\ J/mol$.
The kinetic energy per electron $K=\frac{K_{mol}}{N_A}=\frac{710988}{6.022\times 10^{23}}\ J = 1.181\times 10^{-18}\ J$.

Step3: Calculate the speed of the ejected electron

The kinetic - energy formula is $K=\frac{1}{2}mv^{2}$, where the mass of an electron $m = 9.109\times 10^{-31}\ kg$.
$v=\sqrt{\frac{2K}{m}}$.
$v=\sqrt{\frac{2\times1.181\times 10^{-18}}{9.109\times 10^{-31}}}\ m/s$.
$v=\sqrt{2.6\times10^{12}}\ m/s = 1.61\times 10^{6}\ m/s$.

Answer:

$1.61\times 10^{6}$