Problem 8.2.27^∗

Solution

In order to create a stopping potential for electrons, the potential at the radioactive plate (shown as a thick black line) must be higher than that at the housing (shown as a thick gray line) so that the electric field points away from the plate. Let this potential difference be $V$. Since electrons leaving the plate have energy $eV_0$, their speed $v$ can be determined from

$$\frac{1}{2}mv^2=eV_0\qquad\Rightarrow\qquad v=\sqrt{\frac{2eV_0}{m}},$$

where $m$ is the mass of an electron. Suppose the velocity of an emitted electron makes an angle $\theta$ with a perpendicular line segment between the plate and the housing. Then, it can reach the housing if the kinetic energy due to the component of the velocity along this perpendicular line segment is at least $eV$, which is the potential energy that it will gain. Thus,

$$\frac{1}{2}m(v\cos\theta)^2\ge eV\qquad\Rightarrow\qquad\cos\theta\ge\sqrt{\frac{V}{V_0}}.$$

Since the ratio of the area of the spherical cap through which electrons emitted from a certain point can reach the housing to the the area of the hemisphere through which electrons from that point are emitted is $1-\cos\theta$. The ratio of conventional currents from the housing to the plate when the potential difference $V$ is present or absent is

$$\frac{I}{I_0}=1-\cos\theta\qquad\Rightarrow\qquad I=I_0\left(1-\sqrt{\frac{V}{V_0}}\right).$$