Answer
The maximum possible kinetic energy of the emitted electrons is $9.328\times10^{-19}J/electron$.
Work Step by Step
*Strategy:
We know that if we irradiate a metal with a radiation having photons of total energy $E$:
- A minimum energy of photon ($E_{min}$) that could surpass the 'work function' to overcome the attractive forces holding electrons in the metal would be used to eject the electrons.
- The excess energy of photon would be totally converted into kinetic energy ($K_{E,max}$) of the emitted electrons if there is no loss of excess energy in other forms (heat, etc.).
- Therefore, we could have an equation as follows: $$E=E_{min}+K_{E,max}$$
*Step 1: Find the total energy $E$ of photons of light irradiated on the metal.
- Wavelength of radiation: $\lambda=120nm=1.2\times10^{-7}m$
- Planck's constant: $h\approx6.626\times10^{-34}J.s$
- Speed of light in a vacuum: $c\approx2.998\times10^8m/s$
The total energy of photons of light irradiated on the metal is:
$E=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{1.2\times10^{-7}}\approx1.655\times10^{-18}J$
*Step 2: Find the maximum possible kinetic energy of the emitted electrons.
We would have the maximum kinetic energy when there is no loss of excess energy in other forms.
And we already find out in part a) that $E_{min}\approx7.222\times10^{-19}J$
So,
$K_{E, max}=E - E_{min}= (1.655\times10^{-18}) - (7.222\times10^{-19})=9.328\times10^{-19}J/electron$
