Question

A proton and a singly charged ion of mass 91 atomic mass units (amu) are accelerated through the same potential difference and enter a region of uniform magnetic field moving perpendicular to the magnetic field. What is the ratio of their kinetic energies?  If the radius of the proton's circular path in the magnetic field is 16.8 cm , what is the radius of the path of the singly charged ion of 91 amu?  By what factor must the magnetic field be multiplied by for the singly charge ion to have the same radius path as the proton?

Answer 1

Here, We know that K.E. = 0.5 * m * v^2 = q V

where q is charge on atom , m is mass of atom, v is velocity of atom and V is pot. through which atom is accelerated.

Now, For same charge and Same Pot. Diff.

..................................................................................(1)

Now, Ratio of Kinetic Energy = 1 [Because K.E. = q V which is same for both proton and Singly charged ion]

Then, We know that radius of ion in Magnetic Field =

Here, q and B are same so radius depends upon m v = p = momentum

R_p / R_S = m_p v_p / m_S v_S ............................................................(2)

Now, R_p is radius of proton = 16.8 cm

R_S is the radius of Singly charged ion.

m_p is mass of proton = 1 amu

m_S is mass of Charged ion = 91 amu

v_p is velocity of proton.

v_S is velocity of charged ion.

From (1) and (2), we have

This implies R_S = R_p * (91)^0.5 = 160.21 cm

For,  R_S = R_p ,

B should be multiplied by (91)^0.5 = 9.54 [ Because ]

Now, new B must be 9.54 times B.

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