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Imagine shining sunlight on a metal surface and seeing tiny particles of matter, known as electrons, escape from that surface. This phenomenon is called the photoelectric effect. It presents a fascinating puzzle in physics: light, long considered a wave, appears to act like a stream of particles, transferring energy one particle at a time.
Historically, classical wave theory predicted that increasing the brightness (intensity) of light should gradually increase the energy of emitted electrons. But experiments showed otherwise: electrons were only emitted when light had a certain minimum frequency - regardless of intensity. Moreover, the emission happened instantaneously, defying classical expectations.
This led to the revolutionary idea that light consists of discrete packets of energy called photons. Each photon carries energy proportional to its frequency. This new understanding not only explained the photoelectric effect but also laid the groundwork for quantum physics.
To understand the photoelectric effect quantitatively, Albert Einstein introduced an energy balance equation based on the photon model of light.
When a photon of frequency \( u \) strikes the metal surface, it transfers its entire energy \( h u \) (where \( h \) is Planck's constant) to an electron. Some energy, called the work function (denoted \( \phi \)), is needed to liberate the electron from the metal. Any surplus energy becomes the maximum kinetic energy (\( K_{max} \)) of the emitted electron.
Einstein's Photoelectric Equation:
Why does this matter? It tells us that there is a minimum frequency called the threshold frequency \( u_0 \) below which no electrons are emitted, regardless of light intensity:
This equation and the concept of photons fundamentally changed our understanding of light and energy.
To study the photoelectric effect, physicists use a special apparatus called a photoelectric cell. It typically contains two metal electrodes placed inside a vacuum tube: a metal cathode and a metal anode.
Light of known frequency shines on the cathode, which causes electrons to be emitted due to the photoelectric effect. These electrons are attracted to the anode, creating a measurable electric current. The amount of current and applied voltages help us understand the energy and behaviour of the emitted electrons.
The stopping potential \( V_0 \) is the minimum voltage applied in reverse to stop photoelectrons from reaching the anode. Measuring \( V_0 \) helps determine the maximum kinetic energy of emitted electrons, since the stopping potential counteracts their motion.
| Observation | Classical Wave Theory Prediction | Actual Experimental Result |
|---|---|---|
| Effect of Light Frequency | Electron energy increases smoothly with intensity but independent of frequency | Electrons emitted only if frequency exceeds threshold \( u_0 \) |
| Effect of Light Intensity | Energy of electrons increases with higher intensity | Number of electrons (current) increases with intensity; kinetic energy unchanged |
| Time Delay in Emission | There should be a measurable delay for electrons to accumulate energy | Electron emission is instantaneous (<10⁻⁹ s) |
Step 1: Convert work function to joules:
\( 1 \, \mathrm{eV} = 1.6 \times 10^{-19} \, \mathrm{J} \Rightarrow \phi = 2.5 \times 1.6 \times 10^{-19} = 4.0 \times 10^{-19} \, \mathrm{J} \)
Step 2: Calculate energy of incident photon:
\( E = h u = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 3.98 \times 10^{-19} \, \mathrm{J} \)
Step 3: Apply Einstein's equation:
\( K_{max} = h u - \phi = 3.98 \times 10^{-19} - 4.0 \times 10^{-19} = -0.02 \times 10^{-19} \, \mathrm{J} \)
The kinetic energy is negative, meaning no electrons are emitted because the photon energy is less than the work function.
Answer: No photoelectrons emitted (energy insufficient).
Step 1: Convert work function to joules:
\( \phi = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19} \, \mathrm{J} \)
Step 2: Calculate threshold frequency using \( u_0 = \frac{\phi}{h} \):
\( u_0 = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}} = 5.55 \times 10^{14} \, \mathrm{Hz} \)
Answer: Threshold frequency for sodium is \( 5.55 \times 10^{14} \, \mathrm{Hz} \).
Step 1: Calculate the maximum kinetic energy for both frequencies using \( K_{max} = eV_0 \):
\( K_{max1} = 1.6 \times 10^{-19} \times 0.5 = 8.0 \times 10^{-20} \, \mathrm{J} \)
\( K_{max2} = 1.6 \times 10^{-19} \times 1.1 = 1.76 \times 10^{-19} \, \mathrm{J} \)
Step 2: Use Einstein's equation for each frequency:
\( h u_1 = \phi + K_{max1} \Rightarrow \phi = h u_1 - K_{max1} \)
\( h u_2 = \phi + K_{max2} \Rightarrow \phi = h u_2 - K_{max2} \)
Step 3: Calculate \( h u \) for both frequencies:
\( h u_1 = 6.626 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19} \, \mathrm{J} \)
\( h u_2 = 6.626 \times 10^{-34} \times 7.5 \times 10^{14} = 4.97 \times 10^{-19} \, \mathrm{J} \)
Step 4: Find \( \phi \) values:
\( \phi_1 = 3.98 \times 10^{-19} - 8.0 \times 10^{-20} = 3.18 \times 10^{-19} \, \mathrm{J} \)
\( \phi_2 = 4.97 \times 10^{-19} - 1.76 \times 10^{-19} = 3.21 \times 10^{-19} \, \mathrm{J} \)
The values are close; average \( \phi = 3.2 \times 10^{-19} \, \mathrm{J} \).
Step 5: Convert \( \phi \) in eV:
\( \phi = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0 \, \mathrm{eV} \)
Answer: Work function of the metal is approximately \( 2.0 \, \mathrm{eV} \).
Step 1: Understand what intensity means: it is the number of photons striking per second.
Step 2: More photons mean more electrons emitted per second, so photoelectric current increases proportionally.
Step 3: The energy per photon does not change with intensity, so maximum kinetic energy and stopping potential remain the same.
Answer: Doubling intensity doubles the photoelectric current but does not affect stopping potential.
Step 1: Use the relation \( K_{max} = eV_0 \); \( e = 1.6 \times 10^{-19} \, \mathrm{C} \), \( V_0 = 1.5 \, \mathrm{V} \).
\( K_{max} = 1.6 \times 10^{-19} \times 1.5 = 2.4 \times 10^{-19} \, \mathrm{J} \)
Step 2: Convert kinetic energy to eV:
\( K_{max} = \frac{2.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.5 \, \mathrm{eV} \)
Answer: Maximum kinetic energy is \( 2.4 \times 10^{-19} \, \mathrm{J} \) or 1.5 eV.
When to use: If problem gives wavelength instead of frequency.
When to use: When stopping potential is provided.
When to use: When combining work function, kinetic energy, and photon energy.
When to use: To quickly answer conceptual questions.
When to use: When estimating feasibility of photoelectric emission in problems.
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