Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10−27 kgkg and a charge of 1.60×10−19 CC . The deuteron travels in a circular path with a radius of 6.90 mmmm in a magnetic field with a magnitude of 3.00 TT .

A) Find the speed of the deuteron
B)Find the time required for it to make 12 of a revolution.
C) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Answer 1

a) v=991616m/s

b) t=2.19*10^-8sec

c)V=10263V

Step-by-step explanation:

Given that a deuteron of mass 3.34*10^-27kg with a charge of 1.6*10-19C and travel a path of radius 6.90mm which is 0.0069m in a 3 tesla magnetic field

a)

We are to find the speed of deuteron

Recall that F= qVB

But we know F is also m*a

Hence

M*a= qVB

We're a is the centripetal acceleration directed inwards

Ie a=V^2/r

So M*V^2/r=qVB

V= qBr/M=1.6*10^-19*3*0.0069/3.34*10^-27

Velocity (V)=991616.8m/s

b)

We are to find the time for 12 revolution

Note before the path traced by the deuteron is a semi circular path

Hence T = distance/velocity.

T= πr/v

t=3.142*0.0069/991616.8

t=2.19*10^-8seconds

c). calculating for the potential difference V

Remember that the kinetic energy must equal the potential energy

So

1/2mv^2=/q/V

Hence V= 1/2mv^2/q

V=1/2*3.34*10^-27*991616^2/1.6*10^-19

V=10263.2V