Chapter 20, Problem 68GP

Solution

To calculate: The ratio of radii of proton and electron circular path.

  $\frac{r_p}{r_e}=42.81$

Given information: A proton and electron have the same kinetic energy upon entering a region of constant magnetic field

Formula used:

  $\frac{m{v}^2}{r}=qvB$ here r is radius and m is mass

Here $\frac{m{v}^2}{r}$ is the centripetal force and $qvB$ is the Lorentz force . Here Lorentz fore is provided by the centripetal force.

Calculation: Given $K_P=K_e$ here K represents kinetic energy, p and e represents proton and electron respectively.

  $\frac{1}{2}{m}_p{v}_p{}^2=\frac{1}{2}{m}_e{v}_e{}^2$

  $v_p=\sqrt{\frac{m_e}{m_p}}{v}_e$

  $v_p=\sqrt{\frac{m_e}{1833m_e}}{v}_e$

  $v_p=\sqrt{\frac{1}{1833}}{v}_e$

  $m_p=1833m_e$ ,as the mass of the proton is 1833 times that of the electron.

  $\frac{r_p}{r_e}=\frac{m_p{v}_p}{m_e{v}_e}$

  $\frac{r_p}{r_e}=\frac{1833m_e{v}_e}{m_e{v}_e\sqrt{1833}}=\sqrt{1833}$

  $\frac{r_p}{r_e}=42.81$