$$
E=2.3times 1.6times 10^{-19}
$$

Finally, we have

$$
boxed{E=3.68times 10^{-19}textrm{ J}}
$$

$$
E=2.3times 1.6times 10^{-19}
$$

Finally, we have

$$
E=3.68times 10^{-19}textrm{ J}
$$

Now, the kinetic energy by definition is

$$
E=frac{mv^2}{2}
$$

from where one can express the speed as

$$
v=sqrt{frac{2E}{m}}=sqrt{frac{2times 3.68times 10^{-19}}{9.11times 10^{-31}}}
$$

Finally, we get that

$$
boxed{v=9times 10^5 ,frac{textrm{m}}{textrm{s}}}
$$

$$
K_{max}=-qV_0
$$

Now, we can transfer our values from Joules to eVs but we can more elegantly note that eV is a unit charge times one Volt which gives that in our case

$$
K_{max}=-(-1e)times 5.7V
$$

Finally

$$
boxed{K_{max}=5.7textrm{ eV}}
$$

$$
K_{max}=-qV_0=1.6times 10^{-19}times 3.2
$$

Finally, we get that

$$
boxed{K_{max}=5.12times 10^{-19}textrm{ J}}
$$

$$
W=hf_0=6.63times 10^{-34}times 9.7times 10^{-14}times 6.24 times 10^{18}
$$

Finally

$$
boxed{W=4textrm{ eV}}
$$

$$
W=4textrm{ eV}
$$

$$
K_{max}=E_{ph}-W=frac{hc}{lambda}-hf_0
$$

After we plug in the values we have that

$$
K_{max}=frac{1240times 10^{-9}}{425times 10^{-9}}-1.96=2.92-1.96
$$

Finally, we find that

$$
boxed{K_{max}=0.96textrm{ eV}}
$$

$$
K=E_{ph}-W=frac{hc}{lambda}-W
$$

Now, we can express the work function as

$$
W=frac{hc}{lambda}-K=frac{1240times 10^{-9}}{193}-3.5=6.4-3.5
$$

Finally, we have that

$$
boxed{W=2.9textrm{ eV}}
$$

$$
W=frac{hc}{lambda_0}
$$

so we have that the threshold wavelength is given as

$$
lambda_0=frac{hc}{W}=frac{1240times 10^{-9}}{4.5}
$$

Finally, we have that

$$
boxed{lambda_0=276times 10^{-9}textrm{ m}}
$$

$$
f_{peak}propto T
$$

and the energy is proportional to the temperature

$$
Epropto T^4
$$

So both will increase with the temperature.

$$
E_{ph}=K+W
$$

So the work function is

$$
W=E_{ph}-K=frac{hc}{lambda}-etimes V_{stop}
$$

$$
W=frac{1.24times 10^{-6}}{0.532times 10^{-6}}-1times 1.44
$$

Finally, we have that

$$
boxed{W=0.89textrm{ eV}}
$$

$$
E=frac{hc}{lambda}=frac{1.24times 10^{-6}}{0.65times 10^{-6}}
$$

Finally, we have that

$$
boxed{E=1.91textrm{ eV}}
$$

$$
E=frac{hc}{lambda}=frac{1.24times 10^{-6}}{0.00002times 10^{-6}}
$$

Finally, we have that

$$
boxed{E=62times 10^3textrm{ eV}}
$$

$$
v_1=frac{m_1-m_2}{m_1+m_2}u_1
$$

if we take that the initial velocity of the heavier body is zero. So we see that if

$$
m_2gg m_1
$$

the velocity of the lighter object is approximately the same. So we can conclude that the heavier second object is less kinetic energy will have after the collision so the answer to both questions is no i.e. proton will take less energy from the photon and photon will lose less energy when colliding with the proton.