\({K_{\max }} = hv - {\phi _0} = \) \(hv - h{v_o} = \frac{{hc}}{\lambda } - \frac{{hc}}{{{\lambda _0}}}\)

where \(\lambda_{0}=\) threshold wavelength

or \(\frac{1}{2} m v^{2}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)

Here, \(h=4.14 \times 10^{-15}\,eV\) \(s, c=3 \times 10^{8} \mathrm{ms}^{-1}\)

\({\lambda _o} = 3250 \times {10^{ - 10}}\,\,{\text{m}} = 3250\,\,\mathop A\limits^o \)

\(\lambda  = 2536 \times {10^{ - 10}}\,\,{\text{m}}\, = 2536\mathop A\limits^o ,\)

\(m = 9.1 \times {10^{ - 31}}{\text{kg}}\)

\(hc = 4.14 \times {10^{ - 15}}{\text{eV}}\,\,{\text{s}} \times 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}\)

\( = 12420\,\,{\text{eV}}\mathop A\limits^o \) 

\(\therefore \quad \frac{1}{2}m{v^2} = \) \(12420\left[ {\frac{1}{{2536}} - \frac{1}{{3250}}} \right]{\text{eV}}\)

\(=1.076\, \mathrm{eV}\)

\(v^{2}=\frac{2.152 \mathrm{eV}}{m}=\frac{2.152 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\)

\(\therefore \quad v \approx 6 \times 10^{5} \mathrm{m} \mathrm{s}^{-1}=0.6 \times 10^{6} \mathrm{m} \mathrm{s}^{-1}\)