91Ó°ÊÓ

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

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:

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

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

$PE=−qV$

$i=qN$

$E=NqV\text{ }\mathrm{Δ}t$

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:

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

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

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

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,

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

Using this value in the equation of current, we get

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

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

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

$\begin{array}{c}E=NqV\mathrm{Δ}t\\ =iV\mathrm{Δ}t\\ =(2.27×{10}^{−2}\text{ }A)\left({10}^5\text{ }V\right)\left(3600\text{ }s\right)\\ =8.17×{10}^6\text{ }J\end{array}$

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