Following results are observed when sodium metal is irradiated with different wavelengths. Calculate threshold wavelength and Planck’s constant.

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If the threshold wavelength is $\lambda_{0} n m\left(=\lambda_{0} \times 10^{-9} \mathrm{~m}\right)$ , the K.E. of the radiation would be:
$h\left(\nu-\nu_{0}\right)=\frac{1}{2} m \nu^{2}$
Three equations can be formed by these values:
$\begin{array}{l} h\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m \nu^{2} h c\left(\frac{1}{500 \times 10^{9}}-\frac{1}{\lambda_{0} \times 10^{-9} m}\right)=\frac{1}{2} m\left(2.55 \times 10^{+5} \times 10^{-2} \mathrm{~ms}^{-1}\right) \\ \frac{h c}{10^{-9} m}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(2.55 \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2} \end{array}$
Similarly,
$\begin{array}{l} \frac{h c}{10^{-9} m}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(3.45 \times 10^{+3} m s^{-1}\right)^{2} \\ \frac{h c}{10^{-9} m}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(5.35 \times 10^{+3} \mathrm{~ms}^{-1}\right)^{2} \end{array}$
Dividing equation (3) by equation (1):
$\begin{array}{l} \frac{\left[\frac{\lambda_{0}-400}{400 \lambda_{0}}\right]}{\left[\frac{\left(0.35 \times 10^{+3} m s^{-1}\right)^{2}}{500 \lambda_{0}}\right]}=\frac{5 \lambda_{0}-2000}{\left(2.55 \times 10^{+3} m s^{-1}\right)^{2}} \quad \frac{55.35)^{2}}{4 \lambda_{0}-2000}=\frac{28.6225}{6.5025} \\ \frac{5 \lambda_{0}-2000}{4 \lambda_{0}-2000}=4.40177 \\ 17.6070 \lambda_{0}-5 \lambda_{0}=8803.537-2000 \\ \lambda_{0}=\frac{6805.537}{12.607} \\ \lambda_{0}=539.8 n m \end{array}$