Sodium has a work function of 2 eV. Calculate the maximum energy and speed of the emitted electrons, when sodium is illuminated by radiation of wavelength 150 nm:

The correct option is:
$13.2\times 10{\mathrm{}}^{-19}\text{J and}1.5\times 10{\mathrm{}}^6\text{m/s}$

Step-by-Step Explanation:

1. Identify the Given Parameters:
Work function of sodium, $\Phi =2\text{eV}$
Wavelength of incident radiation, $\lambda =150\text{nm}=150\times 10{\mathrm{}}^{-9}\text{m}$
Planck's constant, $h\approx 6.63\times 10{\mathrm{}}^{-34}\text{J·s}$
Speed of light, $c=3\times 10{\mathrm{}}^8\text{m/s}$
Mass of an electron, $m\approx 9.1\times 10{\mathrm{}}^{-31}\text{kg}$

2. Convert the Work Function to Joules:
Since 1 eV = $1.6\times 10{\mathrm{}}^{-19}\text{J}$:
$\Phi =2\times 1.6\times 10{\mathrm{}}^{-19}\text{J}=3.2\times 10{\mathrm{}}^{-19}\text{J}$

3. Calculate the Incident Photon Energy (E):
Using the photon energy formula:
$E=\frac{hc}{\lambda }$
Substituting the values:
$E=\frac{(6.63\times 10{\mathrm{}}^{-34}\text{J·s})\times (3\times 10{\mathrm{}}^8\text{m/s})}{150\times 10{\mathrm{}}^{-9}\text{m}}$
$E=\frac{1.989\times 10{\mathrm{}}^{-25}}{1.5\times 10{\mathrm{}}^{-7}}\approx 13.26\times 10{\mathrm{}}^{-19}\text{J}$
This corresponds to the incident energy value shown in the option: $13.2\times 10{\mathrm{}}^{-19}\text{J}$.

4. Calculate the Maximum Kinetic Energy (K_max) of Emitted Electrons:
Using Einstein's photoelectric equation:
$K_{\text{max}}=E-\Phi$

5. Calculate the Maximum Speed (v_max):
The relation between kinetic energy and speed is:
$K_{\text{max}}=\frac{1}{2}m{v}_{\text{max}}^2$
Solving for speed:
$v_{\text{max}}=\sqrt{\frac{2K_{\text{max}}}{m}}$
Substitute the values into the equation:
$v_{\text{max}}=\sqrt{\frac{2\times 10.0\times 10{\mathrm{}}^{-19}\text{J}}{9.1\times 10{\mathrm{}}^{-31}\text{kg}}}$