Answer
The stopping potential increases by $~~1.24~V$
Work Step by Step
We can find the maximum kinetic energy of the electron when $\lambda = 250~nm$:
$K_{max} = \frac{h~c}{\lambda} - E_0$
$K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{250\times 10^{-9}~m} - 4.28~eV$
$K_{max} = (7.9512\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J}) - 4.28~eV$
$K_{max} = 4.97~eV - 4.28~eV$
$K_{max} = 0.69~eV$
Therefore, the stopping potential is $~~0.69~V$
We can find the maximum kinetic energy of the electron when $\lambda = 200~nm$:
$K_{max} = \frac{h~c}{\lambda} - E_0$
$K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{200\times 10^{-9}~m} - 4.28~eV$
$K_{max} = (9.939\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J}) - 4.28~eV$
$K_{max} = 6.21~eV - 4.28~eV$
$K_{max} = 1.93~eV$
Therefore, the stopping potential is $~~1.93~V$
The stopping potential increases by $~~1.24~V,~~$ from $0.69~V$ to $1.93~V$
