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lesya [120]
3 years ago
12

I need help plzzzz!!!!!!!!!!!!!!!!

Physics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

You were a freeloader of my questions, so I'll be one too.

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(A) v=3.1\times 10^7 m.s^{-1}

(B) r=0.1903 m

Explanation:

Given:

  • initial energy of hydride ion, KE_i=5 MeV
  • final energy of hydride ion, KE_i=20 MeV
  • mass of ion (=mass of proton, we know), m=1.67\times 10^{-27} kg
  • magnetic field, B=1.7 T

(A)

kinetic energy of 5 MeV ion:

KE_i=5 MeV

\frac{1}{2} m.v^2=5\times 10^{6}\times 1.6\times 10^{-19}

\frac{1}{2} 1.67\times 10^{-27}.v^2=5\times 10^{6}\times 1.6\times 10^{-19}

v=3.1\times 10^7 m.s^{-1}

(B)

Radius of ion when having energy 5.0 MeV:

We know:

F=q.v.B..................................(1)

also,

F=m.\frac{v^2}{r}..........................(2)

where:

F= force on the charge

q= quantity of charge on the particle (an electron in this case)

m= mass of the charged particle

B= magnetic field in which the charge is projected

r= radius of the path circulation of the charge

v= velocity of the charge at the instant

From eq. (1) & (2)

m.\frac{v^2}{r}=q.v.B

r=\frac{m.v}{q.B}

putting the reaspective values:

r=\frac{1.67\times 10^{-27}\times 3.1\times 10^7}{1.6\times 10^{-19}\times 1.7}

r=0.1903 m

8 0
3 years ago
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