An electron with a kinetic energy of 2 MeV is produced in a betatron.Its momentum is approximately A)8.2 $\times$ 10^-3 kg · m/s B)1.31 $\times$ 10^-27 kg · m/s C)7.63 $\times$ 10^-22 kg · m/s D)1.31 $\times$ 10^-21 kg · m/s E)7.63 $\times$ 10^-25 kg · m/s

Question

Quizplus · Accepted Answer

The momentum of a particle can be calculated as p = sqrt(2mK), where m is the mass of the particle and K is its kinetic energy. For an electron, m can be taken as the rest mass, which is 9.11 x 10^(-31) kg. Substituting K = 2 MeV = 2 x 10^6 eV = 2 x 10^6 x 1.6 x 10^(-19) J and m in the above formula, we get p = sqrt(2 x 9.11 x 10^(-31) x 2 x 10^6 x 1.6 x 10^(-19)) = 1.31 x 10^(-21) kg·m/s, which is closest to option D.

An Electron with a Kinetic Energy of 2 MeV Is ×\times×

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