Question regarding radius of circular paths (chapter the nucleus)

[Sanosuke Sagara]

I have my doubt,solution and question in the attachment that followed.Thanks for anybody that spend some time on this question.

 

[Dr.Brain]

OK the problem with your solution is that You have made it a bit too complicated.
First of all ,
After accelerating through a potential , the electrons gain certain amount of K.E , which can be calculate din the following way:

eV= \frac{1}{2}mv^2

From The KE , calculate velocity gained by the electron beam.

Now if you have studied how charged particles behave in magnetic fields, you should know that electrons when enter prependicular to a magnetic field, start moving in a circle.

Force due to a Magnetic field on an electron provides the centripedal force necessary for moving in circle. therefore,

<br /> F= -e(VxB) = m \frac{v^2}{R}<br />

From here calculate the radius of the beam...easy ..isnt it?

 

[Sanosuke Sagara]

Yes,Dr Brain.Thanks for your help and I really appreciate it.Thanks again for your explanation in detail.