Answer
(a) $E_0 = 4.65~eV$
We can assume that the cathode is made of copper.
(b) The stopping potential is $~~1.56~V$
Work Step by Step
(a) We can find the kinetic energy of the electron:
$K = (1.56~V)(1.6\times 10^{-19}~J)$
$K = 2.5\times 10^{-19}~J$
We can find the work function of the metal:
$K_{max} = \frac{h~c}{\lambda} - E_0$
$E_0 = \frac{h~c}{\lambda} - K_{max}$
$E_0 = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{200\times 10^{-9}~m} - 2.5\times 10^{-19}~J$
$E_0 = 7.44\times 10^{-19}~J$
$E_0 = (7.44\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_0 = 4.65~eV$
This value is close to the work function of copper.
We can assume that the cathode is made of copper.
(b) Even if the intensity of the light is doubled, the energy of each photon remains unchanged.
Therefore, the stopping potential is still $~~1.56~V$
