#### Answer

1.45 eV.

#### Work Step by Step

The threshold frequency is $f=\frac{c}{\lambda}=1.10\times10^{15}Hz $.
Now find the work function. It is the energy of this threshold photon.
$\phi =(6.626\times10^{-34}J \cdot s)( 1.10\times10^{15}Hz)=7.29\times10^{-19}J$
Find the energy of the uv photon.
$hf=(6.626\times10^{-34}J \cdot s)( 1.45\times10^{15}Hz)=9.61\times10^{-19}J$
The maximum KE is the uv photon’s energy, minus the work function.
$$K_{max}=hf-\phi=2.32\times10^{-19}J =1.45 eV$$