Answer
$v = 6.8\times 10^5~m/s$
Work Step by Step
In part (a), we found that the stopping potential is $~~1.3~V$
We can find $K_{max}$:
$K_{max} = (1~e)(1.3~V)$
$K_{max} = 1.3~eV$
$K_{max} = (1.3~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$K_{max} = 2.1\times 10^{-19}~J$
We can find the maximum speed of the ejected electrons:
$K_{max} = \frac{1}{2}mv^2 = 2.1\times 10^{-19}~J$
$v^2 = \frac{(2)(2.1\times 10^{-19}~J)}{m}$
$v = \sqrt{\frac{(2)(2.1\times 10^{-19}~J)}{m}}$
$v = \sqrt{\frac{(2)(2.1\times 10^{-19}~J)}{9.109\times 10^{-31}~kg}}$
$v = 6.8\times 10^5~m/s$
