\textbf{176.} The work functions of metals A and B are $4.5\text{ eV}$ and $3\text{ eV}$ respectively. If light with wavelength $150\text{ nm}$ shines on both metals, the photoelectric kinetic energy of metal A's electrons is what percent less than the photoelectric kinetic energy of metal B's electrons?
$$\left(c = 3\times10^{8}\ \frac{\text{m}}{\text{s}},\ h = 4\times10^{-15}\ \text{eV.s}\right)$$
(1) $30\%$ \quad (2) $40\%$ \quad (3) $60\%$ \quad (4) $70\%$