$T = 5800$ K

$textbf{Find:} f_{peak}$

We can compute for $f_{peak}$ using Wien’s displacement law as follows:

$$
begin{align*}
f_{peak} &= (5.88 times 10^{10} text{ Hz}) times 5800 text{ K} \
&= boxed{3.4 times 10^{14} text{Hz}}
end{align*}
$$

$E = 1.2$ J

$v = 0.82$ Hz

$textbf{Find}$ quantum number, $n$

We can calculate $n$ using the formula for quantized energy such that:

$$
begin{align*}
E &= nhf \
n &= dfrac{E}{hf} \
&= dfrac{1.2 text{ J}}{6.626 times 10^{-34} text{ Js} times 0.82 text{ Hz}} \
&= boxed{2.2 times 10^{33}}
end{align*}
$$

$E_1 = 4.2 times 10^{-19} text{ J}$

$E_2 = 1.9 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) = 3.04 times 10^{-19} text{ J}$

To calculate for the frequencies, we use the following formula:

$$
begin{gather}
E = hf
end{gather}
$$

a) Using Eq (1), we can calculate for $f_1$ as follows:

$$
begin{align*}
f_1 &= dfrac{E}{h} \
&= dfrac{4.2 times 10^{-19} text{ J}}{6.626 times 10^{-34} text{ Js}} \
&= boxed{6.3 times 10^{14} text{ Hz}}
end{align*}
$$

$$
begin{align*}
f_2 &= dfrac{E}{h} \
&= dfrac{3.04 times 10^{-19} text{ J}}{6.626 times 10^{-34} text{ Js}} \
&= boxed{4.6 times 10^{14} text{ Hz}}
end{align*}
$$

b) $4.6 times 10^{14}$ Hz

$f = 9.3 times 10^{14}$ Hz

$W_o = 3.6 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) =5.76 times 10^{-19} text{ J}$

$textbf{Find:} KE_{max}$

To calculate for $KE_{max}$, we use the Einstein’s equation for the photoelectric effect as follows:

$$
begin{align*}
E &= hf – W_o \
KE_{max} &= hf – W_o \
&= (6.626 times 10^{-34} text{ Js} times 9.3 times 10^{14} text{ Hz}) – 5.76 times 10^{-19} text{ J} \
&= boxed{4.02 times 10^{-20} text{ J}}
end{align*}
$$