(a) The current is defined as the amount of charge that strikes the target per unit time:
[tex]I= \frac{Q}{t} [/tex]
where Q is the total charge and t the time.
We know the current, [tex]I=0.50 \mu A=0.5 \cdot 10^{-6}A[/tex], and the time, [tex]t=15 s[/tex], so we can calculate the amount of charge that strikes the target during this time interval:
[tex]Q=It=(0.5 \cdot 10^{-6} A)(15 s)=7.5 \cdot 10^{-6}C[/tex]
We know that each proton has a charge of [tex]q=1.6 \cdot 10^{-19}C[/tex], so if we divide the total charge Q by the charge of one proton q, we find the number of protons that strike the target:
[tex]N= \frac{Q}{q}= \frac{7.5 \cdot 10^{-6} C}{1.6 \cdot 10^{-19}C}=4.7 \cdot 10^{13} [/tex]
(b) The total energy given by the beam of protons to the block of aluminium is equal to the kinetic energy of one proton times the number of protons:
[tex]E=NK=(4.7 \cdot 10^{13})(4.9 \cdot 10^{-12}J)=230.3 J[/tex]
This energy is given as heat to the block of aluminium, and the increase in temperature of the block is related to this energy by
[tex]E=mC_s \Delta T[/tex]
where m=15 g=0.015 kg is the mass of the block, [tex]C_s=900 J/kg^{\circ}C[/tex] is the specific heat capacity of aluminium, and [tex]\Delta T[/tex] is the increase in temperature that we want to find.
If we rearrange the equation, we find [tex]\Delta T[/tex]:
[tex]\Delta T= \frac{E}{mC_s}= \frac{230.3 J}{(900 J/kg^{\circ})(0.015 kg)} =17.0^{\circ}C[/tex]
