A metal surface is exposed to 500nm radiation. The threshold...

$\lambda = 500nm \to v = \dfrac{C}{\lambda} = \dfrac{3 \times {10}^{8}}{500 \times {10}^{- 9}} = 6 \times {10}^{14}Hz$

${v}_{0} = 4 . 3 \times {10}^{14}Hz$

For photo electric effect, $hv = {hv}_{0} + KE$

$KE = hv - {hv}_{0} = h\left(v - {v}_{0}\right) = 6 . 6 \times {10}^{- 34}\left(6 \times {10}^{14} - 4 . 3 \times {10}^{14}\right)$

$= 6 . 6 \times 1 . 7 \times {10}^{- 20}J$

$K . E . = \dfrac{1}{2}{mv}^{2}; v = \sqrt{\dfrac{2 \times K \cdot E}{m}}$

$V = \sqrt{\dfrac{2 \times 6 . 6 \times 1 . 7 \times {10}^{- 20}}{9 . 1 \times {10}^{- 31}}} = \sqrt{24 . 65 \times {10}^{10}} = 5 \times {10}^{5}m / s$