Question

A beam of mixture of alpha particles and protons accelerated through same potential difference before entering into the magnetic field of strength



b. If r2= 5cm and then r1=

Answer 1

Answer:

The value of r1 is 5/√2

Explanation:

According to the problem the r2= 5 cm

Now, when we charge a particle q1 with voltage v then it gains energy

q1v= 1/2 mv^2

now the radius is r1= mv1/q1B

                                  = p/q1B

                                 = √2me1/q1B

                                = √2mq1V/q1B

                               = √m1/q1 . √2v/B

Similarly we can write, r2=  √m2/q1.√2v/B

Now r2/r1=  √m2/q1.√2v/B/ √m1/q1 . √2v/B

                  = √m2/q2 . √q1/m1

                    = √q1/q2 . √m2/m1

The mass of alpha particle is four times of proton particles

the charge of alpha particle is two times of protons particles

Therefore,  r2/r1 = √4 x √1/2

             => √2 r1= r2

             = >r1= 5/√2