MODERN BYU 9

Problem 9: Beta-Decay Source Characteristics

Analysis:

The radioactive source acts as an electron pump. The electrons are emitted with energies uniformly distributed between $W_{min}$ and $W_{max}$.

The current $I$ passing through the circuit creates a potential drop $V = IR$ across the resistor. This potential $V$ acts as a retarding potential, stopping electrons with kinetic energy $K < eV$.

Part (a): Constant Current Source

If the retarding potential $eV$ is less than the minimum emission energy $W_{min}$, all emitted electrons reach the outer shell. The current is simply the total emission rate:

$$ I = n e = \text{constant} $$

This holds true as long as $eV \le W_{min}$, or $e(neR) \le W_{min}$.

Part (b): Voltage Source Behavior

When $R$ increases such that $eV > W_{min}$, the potential begins to stop the lower-energy electrons. Since the energy distribution is uniform, the fraction of electrons reaching the shell decreases linearly with increasing $V$.

The current $I$ becomes a function of $V$:

$$ I(V) = I_{max} \left( \frac{W_{max} – eV}{W_{max} – W_{min}} \right) = n e \left( \frac{W_{max} – eV}{W_{max} – W_{min}} \right) $$

Since $V = IR$, we can substitute $I = V/R$ to find the operating voltage:

$$ \frac{V}{R} = ne \frac{W_{max} – eV}{W_{max} – W_{min}} $$

Rearranging to solve for $V$:

$$ V(W_{max} – W_{min}) = ne R W_{max} – ne^2 R V $$ $$ V (W_{max} – W_{min} + ne^2 R) = ne R W_{max} $$ $$ V = \frac{ne R W_{max}}{(W_{max} – W_{min}) + ne^2 R} $$