Question

Three charge proton deuteron and alpha particle are projected in uniform transverse magnetic field what is the ratio of their radii of circular tracks respectively if their kinetic energy is the same

Answer 1

Answer:

The ratio of their radii rp : rd : rα = 1 : √2 : 1 .

Explanation:

Let the radii of proton, deuteron and alpha particle be “rp”, “rd” & “rα”.

All have the same K.E.  

Since all the three charge particles are moving in a uniform magnetic field in circular track therefore,

centripetal force = magnetic force  

mv² / r = q v B

or, r = mv² / qvB

or, r = mv / qB …… (i)

Now,  

Mass of proton = 1

Charge of proton = 1

Mass of deuteron = 2

Charge of deuteron = 1

Mass of alpha particle = 4

Charge of alpha particle = 2

∴ rp = (1m * v) / (1q * B) …. (i)

∴ rd = (2m * v) / (1q * B) …. (ii)

∴ rα = (4m * v) / (2q * B) …. (iii)

We have,  

K.E. = ½ mv²  

Since K.E. is same for all three therefore, we can get the value of “v” for proton, deuteron and alpha particle as follows:

Proton → v = √[(2 * K.E.) / m] …. (iv)

Deuteron → v = √[(K.E.) / m] …. (v)

Alpha particle → v = √[(K.E.) / 2m] … (vi)

The ratio of their radii are

rp : rd : rα  

substituting the values of v from equations (iv), (v) & (vi) in the equations (i), (ii) & (iii), we get

[1m/(1q*B)]*[√{(2*K.E.)/m}] : [2m/(1q*B)]*[√{(K.E.)/m}] : [4m/(2q*B)]*[ √{(K.E.)/2m}]

Or, [1m/(1q*B)]*[√{(2*K.E.)/m}] : [2m/(1q*B)]* [√{(K.E.)/m}] : [4m/(2q * B)]*[√{(K.E.)/2m}]

On cancelling the similar terms  

Or, √2 : 2 : 2*√(1/2)

Or, 2 : 2√2 : 2 …. [dividing by √2 throughout]

Or, 1 : √2 : 1  

∴ rp : rd : rα :: 1 : √2 : 1