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Chapter 4: Q84PP (page 784)

For a particular experiment, helium ions are to be given a kinetic energy of 3.0 MeV. What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) -3.0 MV; (b) +3.0 MV; (c) +1.5 MV; (d) +1.0 MV.

Short Answer

Expert verified

The voltage of the accelerator is V=1.5 MV. So, the answer is (c).

Step by step solution

01

Step 1:

According to conservation law of energy:

鈭�KE-鈭�U=0Kf-Ki=Ui-Uf

Potential energy is U=qV

02

Step 2:

Kf-Ki=Ui-UfKf-0=qVi-VfKf=q鈭�V3MeV=q鈭�V鈭�V=3MeVq=3MeV2e=1.5MV

The voltage of the accelerator is V=1.5 MV. So, the answer is (c).

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