Answer
(a) Aluminum
(b) Al: $1.03\times 10^{15}Hz$
Ca: $6.93\times 10^{14}Hz$
Work Step by Step
(a) We know that energy is directly proportional to frequency. Aluminum has a large work function, which means large energy. Thus, it needs higher frequency to knock out the electron.
(b) We can find the minimum frequency for aluminum as
$Work \space function=hf_{minimum}$
We plug in the known values to obtain:
$6.848\times 10^{-19}=6.63\times 10^{-34}\times f_{minium}$
$\implies f_{minimum}=1.03\times 10^{15}$ Hz
We can find the minimum frequency for calcium as
$Work \space function=hf_{minimum}$
We plug in the known values to obtain:
$4.592\times 10^{-19}=6.63\times 10^{-34}\times f_{minium}$
$\implies f_{minimum}=6.93\times 10^{14}$ Hz
