Following results are observed when sodium metal is irradiated with different wavelengths. Calculate $(a)$ threshold wavelength and $(b)$ Planck's constant.

$\lambda \ (nm)$	$500, 450, 400$
$v \times 10^{-5} \ (cm \ s^{-1})$	$2.55, 4.35, 5.35$

(A) The kinetic energy of the emitted electrons is given by the Einstein photoelectric equation: $h\nu - h\nu_{0} = \frac{1}{2}mv^{2}$,which can be written as $hc(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}) = \frac{1}{2}mv^{2}$.
Using the data provided for $\lambda = 500 \ nm$ $(v = 2.55 \times 10^{5} \ cm/s = 2.55 \times 10^{3} \ m/s)$ and $\lambda = 400 \ nm$ $(v = 5.35 \times 10^{3} \ m/s)$:
$hc(\frac{1}{500 \times 10^{-9}} - \frac{1}{\lambda_{0}}) = \frac{1}{2}m(2.55 \times 10^{3})^{2}$ $(I)$
$hc(\frac{1}{400 \times 10^{-9}} - \frac{1}{\lambda_{0}}) = \frac{1}{2}m(5.35 \times 10^{3})^{2}$ $(II)$
Dividing $(II)$ by $(I)$ and solving for $\lambda_{0}$ gives $\lambda_{0} \approx 540 \ nm$.
Substituting $\lambda_{0}$ back into equation $(I)$ allows for the calculation of Planck's constant $h$,yielding $h \approx 6.626 \times 10^{-34} \ J \ s$.