In a certain particle accelerator, electrons emerge with an energy of 40.0 MeV (1 MeV = 1.60 × 10^-13 J). The electrons emerge not in a steady stream but rather in pulses at the rate of 250 pulses/s. This corresponds to a time interval between pulses of 4.00 ms (Fig. 27.16). Each pulse has a

Question

In a certain particle accelerator, electrons emerge with an energy of 40.0 MeV $\left(1 MeV =1.60 \times 10^{-13} J \right)$ . The electrons emerge not in a steady stream but rather in pulses at the rate of 250 pulses/s. This corresponds to a time interval between pulses of 4.00 ms (Fig. 27.16). Each pulse has a duration of 200 ns, and the electrons in the pulse constitute a current of 250 mA. The current is zero between pulses.

(A) How many electrons are delivered by the accelerator per pulse?

(B) What is the average current per pulse delivered by the accelerator?

(C) What is the peak power delivered by the electron beam?

(A) How many electrons are delivered by the accelerator per pulse?

(B) What is the average current per pulse delivered by the accelerator?

(C) What is the peak power delivered by the electron beam?

Accepted Answer

We use Equation 27.2 in the form dQ = I dt and integrate to find the charge per pulse. While the pulse is on, the current is constant; thus,

[latex] I \equiv \frac{d Q}{d t} [/latex] (27.2).

[latex] Q_{ ext {pulse }}=I \int d t=I \Delta t=\left(250 imes 10^{-3} A ight)\left(200 imes 10^{-9} s ight) [/latex].

[latex] =5.00 imes 10^{-8} C [/latex].

Dividing this quantity of charge per pulse by the electronic charge gives the number of electrons per pulse:

[latex] ext { Electrons per pulse }=\frac{5.00 imes 10^{-8} C / ext { pulse }}{1.60 imes 10^{-19} C / ext { electron }} [/latex].

[latex] =3.13 imes 10^{11} ext { electrons/pulse } [/latex].

(B) What is the average current per pulse delivered by the accelerator?

Average current is given by Equation 27.1, [latex] I_{ av }=\Delta Q / \Delta t [/latex] Because the time interval between pulses is 4.00 ms, and because we know the charge per pulse from part (A), we obtain

[latex] I_{ av }=\frac{\Delta Q}{\Delta t} [/latex] (27.1).

[latex] I_{ av }=\frac{Q_{ ext {pulse }}}{\Delta t}=\frac{5.00 imes 10^{-8} C }{4.00 imes 10^{-3} s }=12.5 \mu A [/latex].

This represents only 0.005% of the peak current, which is 250 mA.

(C) What is the peak power delivered by the electron beam?

By definition, power is energy delivered per unit time interval. Thus, the peak power is equal to the energy delivered by a pulse divided by the pulse duration:

(1) [latex] P _{ ext {peak }}=\frac{ ext { pulse energy }}{ ext { pulse duration }} [/latex].

[latex] =\frac{\left(3.13 imes 10^{11} ext { electrons } / ext { pulse } ight)(40.0 MeV / ext { electron })}{2.00 imes 10^{-7} s / ext { pulse }} [/latex]

[latex] imes\left(\frac{1.60 imes 10^{-13} J }{1 MeV } ight) [/latex].

[latex] =1.00 imes 10^{7} W =10.0 MW [/latex].

We could also compute this power directly. We assume that each electron has zero energy before being accelerated.
Thus, by definition, each electron must go through a potential difference of 40.0 MV to acquire a final energy of 40.0 MeV. Hence, we have

(2) [latex] P _{ ext {peak }}=I_{ ext {peak }} \Delta V [/latex].

[latex] =\left(250 imes 10^{-3} A ight)\left(40.0 imes 10^{6} V ight) [/latex].

= 10.0 MW

Question 27.9: In a certain particle accelerator, electrons emerge with an ...

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