In an experimental technique for treating deep tumors, unstable positively charged pions (π +, elementary particles with a mass of 2.2510-28kg) penetrate the flesh and disintegrate at the tumor site, releasing energy to kill cancer cells. If pions with a kinetic energy of 8 keV are required and if a velocity selector with an electric field strength of 2.2103 V/m is used, what must be the magnetic field strength?

Question

Physics

Sofia

3 January, 12:36

In an experimental technique for treating deep tumors, unstable positively charged pions (π +, elementary particles with a mass of 2.2510-28kg) penetrate the flesh and disintegrate at the tumor site, releasing energy to kill cancer cells. If pions with a kinetic energy of 8 keV are required and if a velocity selector with an electric field strength of 2.2103 V/m is used, what must be the magnetic field strength?

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Answers (1)

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Simone Arellano · Answer 1

Answer: 7.12*10^-4 T

Explanation:

Given

Mass of particles, m = 2.25*10^-28 kg

Energy of the pions, E = 8 keV

Electric field strength, B = 2.2*10^3 V/m

To start with, we know that

K = 1/2*m*v² if we make v subject of formula, we have

v = √ (2K/m) if we substitute our values, we get

v = √[ (2 * 8 * 1.6*10^-16) / 2.25*10^-28

v = √ (2.56*10^-15 / 2.25*10^-28)

v = √1.137*10^13

v = 3.37*10^6

The velocity selector is

v = E/B, Making B subject of formula, so

B = E/v

B = 2.40x10^3 / 3.37*10^6

B = 7.12*10^-4 T

Thus, the magnetic field strength must be 7.12*10^-4 T

= 6.36x10^-4T