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:

$10.1 \times 10^{-19} J$ and $1.5 \times 10^6 m / s$

The correct answer is Option (2) → $13.2 \times 10^{-19} J$ and $1.5 \times 10^6 m / s$

The energy of the photon (E) is -

$E_{photon}=\frac{hc}{λ}=\frac{6.626×10^{-34}×3×10^8}{150×10^{-9}}J$

$=1.33×10^{-18}J$

$(K.E.)_{max}=hv-\phi_0$

$=1.33×10^{-18}-2×1.6×10^{-19}$

$=1.33×10^{-19}-3.2×10^{-19}$

$=(13.3-3.2)×10^{-19}$

$=10.1×10^{-19}$

$K_{max}=\frac{1}{2}mv^2$

$⇒v=\sqrt{\frac{2K_{max}}{m}}$

$⇒v=\sqrt{\frac{2×10.1×10^{-19}}{9.11×10^{-31}}}$

$=10^{-6}×\sqrt{\frac{20.1}{9.11}}$

$≃1.5×10^{-6}m/s$