This question involves understanding the photoelectric effect and how the photocurrent (plate current I) varies with the wavelength λ of incident light when the anode voltage is fixed. Let's break it down step by step.

Step 1: Recall the Photoelectric Effect Basics
In the photoelectric effect, when light of sufficient frequency (or sufficiently short wavelength) falls on a cathode, electrons are emitted. These electrons are attracted to the anode, creating a photocurrent (I). The key equation is Einstein's photoelectric equation:

$KE=h\nu -\varphi =\frac{hc}{\lambda }-\varphi$

where KE is the maximum kinetic energy of emitted electrons, h is Planck's constant, ν is the frequency of light, c is the speed of light, λ is the wavelength, and ϕ is the work function of the cathode material.

Step 2: Understand the Role of Anode Voltage
The anode voltage is kept fixed. This means the accelerating potential is constant. If the voltage is positive and sufficient, it helps in collecting all emitted electrons, leading to saturation current. However, if the voltage is low or negative, it may retard the electrons.

Step 3: Analyze Variation with Wavelength
As wavelength λ is gradually changed (increased or decreased), note that frequency ν = c/λ. So, increasing λ decreases ν.

• For very short λ (high ν), the energy of photons is high, so electrons are emitted with high kinetic energy. The photocurrent I depends on the number of photons (intensity), but here intensity is not mentioned to change, so we assume it constant.
• As λ increases, ν decreases. There is a threshold wavelength λ₀ = hc/ϕ, below which no photoelectric emission occurs.

Step 4: Current Behavior
For λ > λ₀, no electrons are emitted, so I = 0.
For λ ≤ λ₀, emission starts. The kinetic energy of electrons increases as λ decreases (since KE ∝ 1/λ). However, the number of electrons (and hence current I) depends on the number of photons, which is constant if intensity is fixed. So, for λ ≤ λ₀, I should be constant and non-zero, provided the anode voltage is sufficient to collect all electrons.

But wait: if the anode voltage is fixed and not high enough to collect all electrons for very low KE (i.e., when λ is just below λ₀, KE is small), then some electrons may not reach the anode, reducing I. However, the question says "anode voltage is kept fixed", and it is typically set to a value that ensures saturation current for all λ where emission occurs, unless specified otherwise. In standard problems, if voltage is fixed and positive, it is assumed to be saturation voltage.

Therefore, for λ ≤ λ₀, I is constant (saturation current), and for λ > λ₀, I = 0.

Step 5: Graph Interpretation
So, the graph of I vs λ should be:

• Zero for λ > λ₀
• A constant positive value for λ ≤ λ₀

This corresponds to a step function: I is zero until λ = λ₀, and then jumps to a constant value for smaller λ.

Looking at the options, the correct graph is the one where I is zero for larger λ and constant for smaller λ. It should be discontinuous at λ = λ₀.

Final Answer: The correct option is the graph where current I is zero for wavelength greater than a certain value (threshold λ₀), and suddenly becomes constant for wavelengths less than that value.

Related Topics

Photoelectric Effect: This is the phenomenon where electrons are emitted from a material when light shines on it. Key points include threshold frequency, work function, and Einstein's equation.

Einstein's Photoelectric Equation: $KE=h\nu -\varphi$ , which explains the energy balance.

Saturation Current: The maximum photocurrent achieved when all emitted electrons are collected by the anode.

Formulae

1. Einstein's Photoelectric Equation: $K{E}_{\text{max}}=h\nu -\varphi =\frac{hc}{\lambda }-\varphi$

2. Threshold Wavelength: ${\lambda }_0=\frac{hc}{\varphi }$

3. Photocurrent I is proportional to the number of photons incident per second (if intensity is constant), and is constant for λ ≤ λ₀ when saturation conditions are met.