Given V0 = 30 kV = 30 × 103 V
vmax = ?
λmax = ?
1. Since kmax = eV0
So hvmax = eV0
or
vmax = \(\frac{eV_0}{h}\)
= \(\frac{1.6\times10^{-19}\times30\times10^3}{6.62\times 10^{-34}}\)
= 7.24 × 1018 Hz
2. λmin = \(\frac{c}{λ_{max}}\) = \(\frac{3\times10^8}{7.25\times10^{18}}\)
= 0.041 × 10-9
or λmin = 0.041 nm