Question

When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their path directly to the center of the detector, as in the drawing. For the same accelerating potential difference, what magnetic field should be used to send beryllium-10 ions (m = 16.63 × 10-27 kg) to the same location in the detector? Both types of ions are singly ionized (q = +e).

Answer 1

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

$F = qvB$

Now this causes a centripetal acceleration

$F = = mv^2/r$

 so

$qvB = mv^2/r$ ...........(i)

$B = mv/(rq)$  ...............(ii)

If  accelerating potential V is  same and  then  kinetic energy equals the potential energy difference

$\frac{1}{2} mv^2 = Vq$

$v = \sqrt{(2Vq/m)}$      put these value in equation (ii)

$B = m\frac{\sqrt{(2Vq/m)} }{rq}$  

simplifying we get  

$B =m \frac{(\sqrt{ 2Vm/q})}{r}$

for same location r will be same in both case

$B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}$      ..............(iii)

$B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}$    ..........(iv)

 dividing (iv) and (iii) equation we get

$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

${B_{10}} = B_{7} \sqrt{\frac{m_{10}}{{m_7}} }$

$B_{10} = 0.2574T\sqrt{\frac{ (1.663x10^-26}{(1.165x10^-26)}$

so on solving we get  

             =0.3075 T

Answer 2

The required magnetic field should be 0.245 T.

When a particle with charge q enters a constant magnetic field, a magnetic force is generated perpendicular to both the velocity of the particle (v) and the magnetic field (B) in that region, mathematically represented as:

    F = qv × B

this forces the particle to move in a curved path which means it experiences a centripetal force:

$mv^{2} /R = qvB$

⇒ R = mv/qB

According to the question, the radius r of both the ions B(7) and B(10) must be same.

Also, if a charge is a region with potential difference V, the kinetic energy of the particle is:

$\frac{1}{2}mv^{2}=Vq$

Hence, velocity (v) = $\sqrt[]{\frac{2Vq}{m} }$

Now, we know that:

              $R_{7} =R_{10}$

⇒        $\frac{m_{7}\sqrt[]{\frac{2Vq}{m_{7} } } }{qB_{7} } = \frac{m_{10}\sqrt[]{\frac{2Vq}{m_{10} } } }{qB_{10} }$

⇒         $\frac{B_{7} }{B_{10} } = \sqrt {\frac{m_{7} }{m_{10} } }$

⇒           ${B_{10} }={B_{7} }\sqrt {\frac{m_{10} }{m_{7} } }$

      B(10) = 0.205 $\sqrt{\frac{16.63}{11.65} }$ T

      B(10) = 0.245 T is the strength of magnetic field required.

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