Q. 405.0( 1 Vote )

A narrow beam of singly charged carbon ions, moving at a constant velocity of 6.0 × 104ms–1, is sent perpendicularly in a rectangular region having uniform magnetic field B = 0.5 T figure. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10–27) kg, where A is the mass number.



Answer :


Given-



Velocity of a narrow beam of singly-charged carbon ions,


v = 6.0 × 104 m s−1



Strength of magnetic field B = 0.5 T



Separation between the two beams from the incident beam are 3.0 cm and 3.5 cm.



Mass of an ion = A(1.6 × 10−27) kg



The radius of the curved path taken by the beam 1 -




The radius of the curved path taken by the beam 2-



The radius of the circle is given by,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C


For the first beam




where


m1 = mass of the first isotope


q = s the charge.


For the second beam




where


m2 = mass of the second isotope


q = the charge.



from (1) and (2),




Now, we know,








Also, from (3)





Looking at the mass of the material obtained, we can infer that these are the two isotopes of carbon used are 12C6 and 14C6.


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