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		+	### Statement
		+
		+	$8.1.13.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Problem
		+
		+	$8.1.13.$ In a vacuum diode, there are two electrodes: a cathode from which electrons "evaporate" (it is specially heated), and an anode, which is hit by electrons emitted from the cathode. Explain why the diode can be used as a rectifier. The figure shows how, at a constant voltage between the anode and cathode, the current in the anode circuit depends on the cathode temperature. Explain this dependence qualitatively.
		+
		+	### Solution
		+
		+	#### 1. Why the diode can be used as a rectifier
		+
		+	The operation of a vacuum diode is based on the phenomenon of thermionic emission. The cathode is specially heated, and electrons, having received sufficient thermal energy to overcome the work function, leave its surface. The anode remains cold and does not emit electrons.
		+
		+	If a forward voltage is applied to the diode (anode is connected to the positive terminal, cathode to the negative), the electric field will accelerate the emitted electrons towards the anode, and a current will flow in the circuit. If a reverse voltage is applied (anode to negative, cathode to positive), the field will repel the electrons back to the cathode. Since the cold anode is not a source of electrons, the reverse current will be zero. Due to this strict one-way conductivity, the vacuum diode serves as a rectifier.
		+
		+	#### 2. Qualitative explanation of the $I(T)$ dependence at $U = \text{const}$
		+
		+	The graph of the anode current versus the cathode temperature has two distinct regions:
		+
		+	- Rapid growth region (emission-limited regime): At relatively low temperatures, the cathode emits few electrons, and the applied voltage $U$ is sufficient to transfer all of them to the anode. In this case, the current is limited only by the cathode's ability to emit electrons. This dependence is described by the Richardson-Dushman equation:
		+	$$ I \propto T^2 \exp\left(-\frac{A_{\text{out}}}{kT}\right) $$
		+	where $A_{\text{out}}$ is the electron work function, and $k$ is the Boltzmann constant. This explains the steep nonlinear rise of the curve.
		+
		+	- Saturation region (space-charge-limited regime): With a further increase in temperature, the cathode begins to emit an excess number of electrons. They accumulate in the vacuum between the electrodes, forming a negatively charged electron cloud (space charge). The electric field of this cloud repels a part of the newly emitted electrons back to the cathode surface.
		+
		+	In this regime, the current through the diode no longer increases with temperature, as it is limited by the ability of the external electric field of the anode to "pull" electrons from this cloud. In this region, the current obeys the "three-halves" law (Child-Langmuir law):
		+	$$ I \propto U^{3/2} $$
		+	Since the voltage $U$ is constant according to the problem statement, the current also remains constant, and the graph reaches a horizontal plateau.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,35 @@
			### Statement

			$8.1.13.$ [Insert the problem statement]

			### Solution

			### Problem

			$8.1.13.$ In a vacuum diode, there are two electrodes: a cathode from which electrons "evaporate" (it is specially heated), and an anode, which is hit by electrons emitted from the cathode. Explain why the diode can be used as a rectifier. The figure shows how, at a constant voltage between the anode and cathode, the current in the anode circuit depends on the cathode temperature. Explain this dependence qualitatively.

			### Solution

			#### 1. Why the diode can be used as a rectifier

			The operation of a vacuum diode is based on the phenomenon of thermionic emission. The cathode is specially heated, and electrons, having received sufficient thermal energy to overcome the work function, leave its surface. The anode remains cold and does not emit electrons.

			If a forward voltage is applied to the diode (anode is connected to the positive terminal, cathode to the negative), the electric field will accelerate the emitted electrons towards the anode, and a current will flow in the circuit. If a reverse voltage is applied (anode to negative, cathode to positive), the field will repel the electrons back to the cathode. Since the cold anode is not a source of electrons, the reverse current will be zero. Due to this strict one-way conductivity, the vacuum diode serves as a rectifier.

			#### 2. Qualitative explanation of the $I(T)$ dependence at $U = \text{const}$

			The graph of the anode current versus the cathode temperature has two distinct regions:

			- Rapid growth region (emission-limited regime): At relatively low temperatures, the cathode emits few electrons, and the applied voltage $U$ is sufficient to transfer all of them to the anode. In this case, the current is limited only by the cathode's ability to emit electrons. This dependence is described by the Richardson-Dushman equation:
			$$ I \propto T^2 \exp\left(-\frac{A_{\text{out}}}{kT}\right) $$
			where $A_{\text{out}}$ is the electron work function, and $k$ is the Boltzmann constant. This explains the steep nonlinear rise of the curve.

			- Saturation region (space-charge-limited regime): With a further increase in temperature, the cathode begins to emit an excess number of electrons. They accumulate in the vacuum between the electrodes, forming a negatively charged electron cloud (space charge). The electric field of this cloud repels a part of the newly emitted electrons back to the cathode surface.

			In this regime, the current through the diode no longer increases with temperature, as it is limited by the ability of the external electric field of the anode to "pull" electrons from this cloud. In this region, the current obeys the "three-halves" law (Child-Langmuir law):
			$$ I \propto U^{3/2} $$
			Since the voltage $U$ is constant according to the problem statement, the current also remains constant, and the graph reaches a horizontal plateau.

			#### Answer

			[Insert a concise answer or boxed result]