(a) Assuming the threshold wavelength to be ${\mathrm{\lambda }}_0 \mathrm{nm}\left(={\mathrm{\lambda }}_0 \mathrm{x} {10}^{-9} \mathrm{m}\right)$, the kinetic energy of the radiation is given as:

$\mathrm{h}\left(\mathrm{v} - {\mathrm{v}}_0\right) = \frac{1}{2} {\mathrm{mv}}^2$

Three different equalities can be formed by the given value as:

$\mathrm{hc}\left(\frac{1}{\mathrm{\lambda }} - \frac{1}{{\mathrm{\lambda }}_0}\right) = \frac{1}{2} {\mathrm{mv}}^2$
$\mathrm{hc}\left(\frac{1}{500 \mathrm{x} {10}^9} - \frac{1}{{\mathrm{\lambda }}_0 \mathrm{x} {10}^{-9} \mathrm{m}}\right) = \frac{1}{2} \mathrm{m}\left(2.55 \mathrm{x} {10}^{+5} \mathrm{x} {10}^{-2} \mathrm{m}/\mathrm{s}\right)$
$\frac{\mathrm{hc}}{{10}^{-9}\mathrm{m}}\left[\frac{1}{500} - \frac{1}{{\mathrm{\lambda }}_0}\right] = \frac{1}{2}\mathrm{m}{\left(2.55 \mathrm{x} {10}^{+3} \mathrm{m}/\mathrm{s}\right)}^2 \left(1\right)$

Similarly,

$\frac{\mathrm{hc}}{{10}^{-9}\mathrm{m}}\left[\frac{1}{450} - \frac{1}{{\mathrm{\lambda }}_0}\right] = \frac{1}{2}\mathrm{m}{\left(3.45 \mathrm{x} {10}^{+3} \mathrm{m}/\mathrm{s}\right)}^2 \left(2\right)$
$\frac{\mathrm{hc}}{{10}^{-9}\mathrm{m}}\left[\frac{1}{400} - \frac{1}{{\mathrm{\lambda }}_0}\right] = \frac{1}{2}\mathrm{m}{\left(5.35 \mathrm{x} {10}^{+3} \mathrm{m}/\mathrm{s}\right)}^2 \left(3\right)$

Dividing equation (3) by equation (1):

$\frac{\left[\frac{1}{400} - \frac{1}{{\mathrm{\lambda }}_0}\right]}{\left[\frac{1}{500} - \frac{1}{{\mathrm{\lambda }}_0}\right]} = \frac{{\left(5.35 \mathrm{x} {10}^{+3} \mathrm{m}/\mathrm{s}\right)}^2}{{\left(2.55 \mathrm{x} {10}^{+3} \mathrm{m}/\mathrm{s}\right)}^2}$
$\frac{5{\mathrm{\lambda }}_0 - 2000}{4{\mathrm{\lambda }}_0 - 2000} = {\left(\frac{5.35}{2.55}\right)}^2 = \frac{28.6225}{6.5025}$
$\frac{5{\mathrm{\lambda }}_0 - 2000}{4{\mathrm{\lambda }}_0 - 2000} = 4.40177$
$17.6070{\mathrm{\lambda }}_0 - 5{\mathrm{\lambda }}_0 = 8803.537 - 2000$
${\mathrm{\lambda }}_0 = \frac{6805.537}{12.607}$
${\mathrm{\lambda }}_0 = 539.8 \mathrm{nm}$
${\mathrm{\lambda }}_0 = 540 \mathrm{nm}$

So, threshold wavelength $\left({\mathrm{\lambda }}_0\right)$ = 540 nm