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