The work-function of a metal is \(1 \mathrm{eV}\). Light of wavelength \(3000 Å\) is incident on this metal surface. The velocity of emitted photoelectrons will be

Given, Work-function, \(\phi_{0}=\mathrm{leV}=1.6 \times 10^{-19} \mathrm{~J}\) Wavelength, \(\lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m}\)
As, we know that,
\(E=h v=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{3000 \times 10^{-10}}\)
\(=6.6 \times 10^{-19} \mathrm{~J}\)
Also, \(E=\phi_{0}+\mathrm{KE}\)
\(\Rightarrow \quad \mathrm{KE}=E-\phi_{0} \Rightarrow \frac{1}{2} m v^{2}=E-\phi_{0}\)
\(v^{2}=\frac{\left(6.6 \times 10^{-19}-1.6 \times 10^{-19}\right) \times 2}{9.1 \times 10^{-31}}\)
\(=\frac{5 \times 10^{-19} \times 2}{9.1 \times 10^{-31}}=1.1 \times 10^{12}\)
\(v=1.01 \times 10^{6}\)
\(\simeq 1 \times 10^{6} \mathrm{~m} / \mathrm{s}\)