Answer
(a) $369 eV$
(b) $4.0\times 10^{14}Hz \le f \lt 1.03\times 10^{15}Hz$
Work Step by Step
(a) We know that
$K_{max}=hf-W_{\circ}$
We plug in the known values to obtain:
$K_{max}=(6.63\times 10^{-34}J.s)(9.0\times 10^{16}Hz)-(4.28eV)$
$K_{max}=369eV$
(b) We know that
$K_{max}=hf-W_{\circ}$
We plug in the known values to obtain:
$0J=(6.63\times 10^{-34}J.s)f-(4.28eV)$
This can be rearranged as:
$f=\frac{[4.28eV(\frac{1.6\times 10^{-19}J}{1eV})]}{6.63\times 10^{-34}J.s}$
$f=1.03\times 10^{15}Hz$
Thus, the range of frequencies for which no electrons are emitted is given as
$4.0\times 10^{14}Hz$ to $1.03\times 10^{15}Hz$
