Atoms

In a circular motion, a body changes its direction by

90%5E%7B%5Ccirc%7D

in one - fourth of its time period. Given, magnetic field,

B%3D2%20mT

Let

T_%7Bp%7D

be the time-period of revolution of proton in the magnetic field.

T_%7Bp%7D%20%3D%5Cfrac%7B2%20%5Cpi%20m_%7Bp%7D%7D%7Be%20B%7D%20%5Cdots

(i)

m_%7B%5Calpha%7D%20%5Csimeq%204%20m%20p

and

q_%7B%5Calpha%7D%3D2%20q%5E%7Bp%7D

and

T_%7B%5Calpha%7D

be the time-period of revolution of

%5Calpha

-particle in the magnetic field,

T_%7B%5Calpha%7D%3D%5Cfrac%7B2%20%5Cpi%20m_%7Bp%7D%7D%7Bq%20B%7D%3D%5Cfrac%7B2%20%5Cpi%5Cleft(4%20m_%7Bp%7D%5Cright)%7D%7B(2%20e)%20B%7D%20%5Cdots

(ii) Now, the ratio of time-period of proton and

%5Calpha

-particle, by dividing E (i) and (ii),

%5Cfrac%7BT_%7Bp%7D%7D%7BT_%7B%5Calpha%7D%7D%20%3D%5Cfrac%7B1%7D%7B2%7D

%5CRightarrow%20T_%7B%5Calpha%7D%3D2%20T_%7Bp%7D

Hence, the time-period of

%5Calpha

-particle is double of the proton, i.e. if proton covers

90%5E%7B%5Ccirc%7D

of angle from its starting, then

%5Calpha

-particle will cover

45%5E%7B%5Ccirc%7D

of the angle.

Chapter Questions
1 MCQs