Respuesta :
Solution :
Given that :
The energy of the protons, K.E. = 1.10 keV
[tex]$= 1.10 \times 10^3 \ eV $[/tex]
The current produced by the generator is I = 5 mA
[tex]$= 5 \times 10^{-3} \ A$[/tex]
Now [tex]$1 \ eV = 1.6 \times 10^{-19 }\ J$[/tex]
Mass of the proton, m = [tex]$1.67 \times 10^_{-27} $[/tex] kg
Charge of the proton, [tex]$q_p = 1.6 \times 10^{-19} \ C$[/tex]
a). Therefore using the formula for K.E. we can find out the velocity of the proton.
[tex]$K.E. =\frac{1}{2}mv^2$[/tex]
[tex]$v=\sqrt{\frac{2K.E.}{m}}$[/tex]
[tex]$v=\sqrt{\frac{2\times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$[/tex]
[tex]$= 4.38 \times 10^5 \ m/s$[/tex]
b). We know that the current is :
[tex]$I=\frac{\Delta Q}{\Delta t}$[/tex]
Therefore, the total charge in one second is given by :
[tex]$\Delta Q = I \times \Delta t$[/tex]
[tex]$= 5 \times 10^{-3} \times 1$[/tex]
[tex]$= 5 \times 10^{-3}\ C$[/tex]
So, the number of protons in this charge is given by :
[tex]$n = \frac{\Delta Q}{q_p}$[/tex]
[tex]$=\frac{5 \times 10^{-3} }{1.6 \times 10^{-19}}$[/tex]
[tex]$= 3.13 \times 10^{16}$[/tex] protons
