Question

A mass spectrometer (Figure) is used to separate uranium ions of mass$\mathbf{3}\mathbf{.92}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}$kg and charge$\mathbf{3}\mathbf{.20}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}$C from related species. The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180° and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup.

(a)What is the magnitude of the (perpendicular) magnetic field in the separator?If the machine is used to separate out 100 mg of material per hour

(b)Calculate the current of the desired ions in the machine.

(c)Calculate the thermal energy produced in the cup in 1.00 h.

Short answer

a) The magnitude of the perpendicular magnetic field in the separator is.$0.495T$

b) The current of the desired ions in the machine is .$2.27\times {10}^{-2}A$

c) The thermal energy produced in the cup in $1.00h$is$8.17\times {10}^{6}J$ .

Step-by-step solution

Listing the given quantities

• Mass of uranium$m=3.92\times {10}^{-25}kg$
• Charge of uranium$q=3.20\times {10}^{-19}C$
• Potential difference$V=100kV={10}^{5}V$
• Radius$r=1.00m$
• Time$t=1.00h=3600s$

Understanding the concept of kinetic energy and magnetic field  

We use the conservation of energy principle in the magnetic field and the radius of the circular path in the magnetic field to find the magnitude of the magnetic field in the separator. We can find the current of the desired ions in the machine using the formula of current for the N number of ions passing per unit time. We use the energy deposits by each ion to find the thermal energy produced in the cup.

Formula:

$r=\frac{mv}{qB}$

$KE=\frac{1}{2}m{v}^2$

$PE=-qV$

$i=qN$

$E=NqV\Delta t$

(a) Calculation of the magnitude of the perpendicular magnetic field in the separator

We have for the radius of the circular orbit,$r=\frac{mv}{qB}$.

But, from the conservation of mechanical energy,

$\begin{array}{c}KE+PE=0\\ \frac{1}{2}m{v}^2-qV=0\\ v=\sqrt{\frac{2qV}{m}}\end{array}$

So,

$\begin{array}{c}r=\frac{m}{qB}\sqrt{\frac{2qV}{m}}\\ =\frac{1}{B}\sqrt{\frac{2mV}{q}}\end{array}$

Now, from figure 28.12, we have:

$\begin{array}{c}x=2r\\ =2.00m\end{array}$

$x=\frac{2}{B}\sqrt{\frac{2mV}{q}}$

$\begin{array}{c}B=\sqrt{\frac{8mV}{q{x}^2}}\\ =\sqrt{\frac{8(3.92\times {10}^{-25}kg)\left({10}^{5}V\right)}{(3.20\times {10}^{-19}C){(2.00m)}^2}}\\ =0.495T\end{array}$

Therefore, the magnitude of the perpendicular magnetic field in the separator is.$0.495T$

(b) Calculation of the current of the desired ions in the machine

The current is given by:

$i=qN$

Where$N$is the number of ions per unit of time.

So, the mass separated per unit time can be expressed as :

$M=mN$

$N=\frac{M}{m}$

So,

$i=q\frac{M}{m}$

Now,

$\begin{array}{c}M=\frac{100mg}{1h}\\ =\frac{100\times {10}^{-6}kg}{3600s}\\ =2.78\times {10}^{-8}\frac{kg}{s}\end{array}$

Using this value in the equation of current, we get

$\begin{array}{c}i=\frac{(3.20\times {10}^{-19}C)(2.78\times {10}^{-8}\frac{kg}{s})}{(3.92\times {10}^{-25}kg)}\\ =2.27\times {10}^{-2}A\end{array}$

Therefore, the current of the desired ions in the machine is.$2.27\times {10}^{-2}A$

(c) Calculation of the thermal energy produced in the cup in 1.00 h

If each ion deposits energy$qV$, then for$N$number of ions, energy deposits in the time interval$\Delta t$is given by:

$\begin{array}{c}E=NqV\Delta t\\ =iV\Delta t\\ =(2.27\times {10}^{-2}A)\left({10}^{5}V\right)\left(3600s\right)\\ =8.17\times {10}^{6}J\end{array}$

Therefore, thethermal energy produced in the cup in $1.00h$is$8.17\times {10}^{6}J$.