Which of the graphs shown below does not represent the relationship between incident light and the e

Q: Why doesn't the kinetic energy of electrons depend on the intensity of light?

Intensity refers to the number of photons hitting the surface per unit area per unit time. Increasing intensity means more photons, which ejects more electrons. However, the energy of each individual photon depends only on its frequency ($E = h\nu$). Since one electron absorbs only one photon, its maximum kinetic energy is determined solely by the photon's frequency, not the total number of photons.

Q: Why is Graph C a step function?

If the incident frequency is below the threshold frequency ($\nu_0$), the photons don't have enough energy to overcome the work function, so zero electrons are emitted. Once the frequency reaches $\nu_0$, emission begins. For any frequency above $\nu_0$, the number of emitted electrons depends on the intensity (number of photons), not the frequency. Assuming constant intensity, the number of electrons jumps from zero to a constant value, creating a step.

Q: What does the slope of the K.E. vs frequency graph represent?

If you look at the equation $K.E. = h\nu - \Phi$ and compare it to $y = mx + c$, the slope '$m$' corresponds to '$h$', which is Planck's constant. This slope is universal and identical for all metals.

Q: Why can't the K.E. vs frequency graph start from the origin?

If the graph started at the origin, it would mean that even light with almost zero frequency could eject an electron with zero kinetic energy. This implies the electrons are completely free and not bound to the metal at all (Work Function = 0). In reality, electrons are bound to the nucleus, so a minimum threshold frequency is required to just free them.

Q: What is the significance of the negative y-intercept in the correct K.E. vs frequency graph?

If you extrapolate the straight line of the K.E. vs frequency graph backwards to intersect the y-axis (where $\nu = 0$), the intercept value is $-\Phi$ (or $-h\nu_0$). The magnitude of this intercept gives you the exact value of the metal's work function.

Question

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

Sigma Percentile · Accepted Answer

### The Elegance of the Photoelectric Effect

The photoelectric effect is one of the most beautiful phenomena in modern physics, serving as the bedrock for quantum mechanics. When light strikes a metal surface, it can eject electrons, but this process doesn't follow classical wave theory. Instead, it obeys Einstein's elegant photoelectric equation:

$$K.E._{max} = h
u - \Phi$$

Here, $K.E._{max}$ is the maximum kinetic energy of the ejected electron, $h
u$ is the energy of the incident photon, and $\Phi$ (or $E_0$) is the work function—the minimum energy required to rip an electron away from the metal's surface.

### Decoding the Graphs

To solve this problem, we must act as detectives, interrogating each graph to see if it aligns with physical reality.

**Graph A: Kinetic Energy vs. Incident Energy**
If we rewrite our equation as $K.E. = E - E_0$, it perfectly mirrors the equation of a straight line, $y = mx + c$. The slope $m$ is $1$, and the x-intercept is $E_0$. Graph A shows exactly this: a straight line starting at a positive energy value. It is a flawless representation.

**Graph B: Kinetic Energy vs. Intensity**
Classical physics predicted that brighter light (higher intensity) would give electrons more energy. Quantum mechanics says no! Intensity merely means *more* photons are hitting the surface per second, not that individual photons are more energetic. Therefore, the maximum kinetic energy of a single ejected electron is completely independent of intensity. Graph B shows a flat, horizontal line, which is absolutely correct.

**Graph C: Number of Electrons vs. Frequency**
Imagine trying to buy a $100 item with a handful of $10 bills. It doesn't matter how many bills you have; no single bill is enough. Similarly, if the incident light's frequency $
u$ is below the threshold frequency $
u_0$, no electrons are ejected, regardless of intensity. Once $
u \ge 
u_0$, electrons are emitted, and their number depends on the intensity of the light. This creates a step function, exactly as shown in Graph C.

### The Fatal Flaw in Graph D

Finally, we arrive at Graph D, which plots Kinetic Energy against Frequency. According to $K.E. = h
u - h
u_0$, this should indeed be a straight line. However, look closely at where the line begins.

Graph D shows the line starting precisely at the origin $(0,0)$. Mathematically, this implies that when $
u = 0$, $K.E. = 0$. Plugging this into our equation yields:

$$0 = h(0) - \Phi \implies \Phi = 0$$

This suggests the metal has a work function of zero! Physically, this is impossible. Every metal binds its electrons with some finite energy. A correct graph must have a positive x-intercept at $
u_0$ (the threshold frequency) and, if extrapolated, a negative y-intercept at $-h
u_0$.

Because Graph D violates this fundamental physical constraint, it is the incorrect representation, making it our final answer.

Animated Solution for Chemistry - Atomic Structure: Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?

Select Answer:

Visualized Solution

K.E._{max} = h\nu - \Phi

K.E. = E - E_0

K.E. \text{ vs Intensity}

\text{Number of } e^- \text{ vs } \nu

K.E. = h\nu - h\nu_0

\Phi = 0 \text{ ?}

\text{Correct Graph}

\text{Conclusion}

The Sigma Insight: Wave Particle Duality

The Elegance of the Photoelectric Effect

Decoding the Graphs

The Fatal Flaw in Graph D

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