Let the threshold wavelength = $\lambda_0$ nm = $\lambda_0 \times 10^{-9} \mathrm{m}$ then $hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$ = $\frac{1}{2} m v^2$ Substituting the given three experiments, $\frac{hc}{10^{-9}}\left(\frac{1}{500} - \frac{1}{\lambda_0}\right)$ = $\frac{1}{2} m (2.55 \times 10^5)^2$ ... (i) $\frac{hc}{10^{-9}}\left(\frac{1}{450} - \frac{1}{\lambda_0}\right)$ = $\frac{1}{2} m (4.35 \times 10^5)^2$ ... (ii) $\frac{hc}{10^{-9}}\left(\frac{1}{400} - \frac{1}{\lambda_0}\right)$ = $\frac{1}{2} m (2.55 \times 10^5)^2$ ... (iii) Dividing equation (ii) by equation (i) we get, $\frac{\lambda_0 - 450}{450\lambda_0}$ × $\frac{500\lambda_0}{\lambda_0 - 500}$ = $\left(\frac{4.35}{2.55}\right)^2$ or $\frac{\lambda_0 - 450}{\lambda_0 - 500}$ = $\frac{450}{500} \times \left(\frac{4.35}{2.55}\right)^2$ or $\frac{\lambda_0 - 450}{\lambda_0 - 500}$ = $\frac{9}{10} \times \frac{18.92}{6.50}$ or $\frac{\lambda_0 - 450}{\lambda_0 - 500}$ = 2.62 or $\lambda_0$ – 450 = 2.62 $\lambda_0$ – 1310 or – 450 + 1310 = 2.62 $\lambda_0 - \lambda_0$ or 860 = 1.62 $\lambda_0$ ⇒ $\lambda_0$ = $\frac{860}{1.62}$ = 530.86 nm ≈ 531 nm Putting this in equation (iii), we get $\frac{h \times 3 \times 10^8}{10^{-9}} \left(\frac{1}{400} - \frac{1}{531}\right)$ = $\frac{1}{2} (9.1 \times 10^{-31}) (5.20 \times 10^5)^2$ h × 3 × $10^{17} \left(\frac{131}{531 \times 400}\right)$ = $\frac{1}{2}$ × 27.04 × $10^{10}$ × 9.1 × $10^{-31}$ ⇒ h = 6.64 × $10^{-34} \mathrm{Js}$.