Answer to Question #137175 in Electric Circuits for Kamogelo

Question

A mass spectrum is used to separate two isotopes of uranium with masses m1 and m2 where m2>m1.the two types of uranium atom exit an ion source S with the same charge of +e and are accelerated through a potential difference V.the charged atoms then enter a constant,uniform magnetic field B .of r1=0,5049 m and r2=0,5081 m.what is the value of the ratio m1/m2

Answer 1

Explanations & Calculations

• Refer to the figure attached
• As the particle escape the accelerating region it attains some velocity & work done within this region on the particle gives the kinetic energy associated with that velocity. Therefore,

$\qquad\qquad \begin{aligned} \small W &= eV=\frac{1}{2}mu^2\\ \small u&= \small \sqrt{\frac{2eV}{m}}\\ \\ &\small \text{for both particles}\\ \therefore u_1&= \small \sqrt{\frac{2eV}{m_1}} \,\,\text{and} \,\,\,\,\, u_2= \small \sqrt{\frac{2eV}{m_2}} \end{aligned}$

• Within the magnetic field active region particle experiences a centripetal force which leads it in a curved path of radius r. Then by the application of Newton's second law towards the center,

$\qquad\qquad \begin{aligned} \small Beu &= \small m\frac{u^2}{r}\\ \small r&= \small \frac{mu}{Be}\\ &= \small \frac{m}{Be}\sqrt{\frac{2eV}{m}}\\ &= \small \sqrt{\frac{2V}{e}} \frac{\sqrt{m}}{B}\\ \small r &\small \propto\sqrt{m} \\ \\ &\small \text{for both particles}\\ \small r_1 & \small \propto\sqrt{m_1}\,\,\,\,\,\text{and} \,\,\,\,r_2 \small \propto \sqrt{m_2} \end{aligned}$

• Therefore,

$\qquad\qquad \begin{aligned} \small \frac{m_1}{m_2} &= \small \Big(\frac{r_1}{r_2} \Big)^2\\ &= \small \Big(\frac{0.5049}{0.5081} \Big)^2\\ &= \small \bold{0.9874} \end{aligned}$