Given W0 = 2.14 eV
= 2.14 × 1.6 × 10-19 J
= 3.424 × 10-19J
v = 6 × 1014 Hz
1. Kmax = hv – W0
= 6.62 × 10-34 × 6 × 1014 – 3.424 × 1019
= 3.972 × 10-19 – 3.424 × 10-19
= 0.54 × 10-19 J.
2. Since eV0 =kmax
∴ v0 = \(\frac{k_{max}}{e}\) = \(\frac{0.54\times10^{-19}}{1.6\times10^{-19}}\)
or v0 = 0.34 V
Since \(\frac{1}{2}\) mV2max = Kmax
∴ V2max = \(\frac{2k_{max}}{m}\)
= \(\frac{2\times0.54\times10^{-19}}{9.1\times10^{-31}}\) = 0.1186
or vmax = 0.344 × 106 ms-1 = 344 kms-1