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Dual Nature of Radiation and Matter - Experimental Study of Photoelectric Effect

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Photoelectric Effect is the phenomenon of emission of electrons from a metal surface when light of a sufficiently high frequency (above the threshold frequency ν0\nu_0) falls on it.

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Experimental study shows that the number of photoelectrons emitted per second (photocurrent) is directly proportional to the intensity of incident radiation, provided the frequency is above ν0\nu_0.

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The maximum kinetic energy KmaxK_{max} of photoelectrons depends linearly on the frequency of the incident radiation and is independent of its intensity.

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Stopping Potential (V0V_0) is the minimum negative (retarding) potential applied to the collector plate for which the photocurrent becomes zero. It is related to maximum kinetic energy by Kmax=eV0K_{max} = eV_0.

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Saturation Current is the maximum value of photocurrent reached when all the photoelectrons emitted from the emitter reach the collector. It increases with the intensity of incident radiation.

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The Threshold Frequency (ν0\nu_0) is the minimum frequency of incident radiation below which no photoelectric emission occurs, regardless of the intensity of light.

📐Formulae

Kmax=12mvmax2=eV0K_{max} = \frac{1}{2}mv_{max}^2 = eV_0

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}

hν=ϕ0+Kmaxh\nu = \phi_0 + K_{max}

eV0=hν−ϕ0eV_0 = h\nu - \phi_0

V0=(he)ν−ϕ0eV_0 = \left( \frac{h}{e} \right) \nu - \frac{\phi_0}{e}

💡Examples

Problem 1:

The work function of cesium metal is ϕ0=2.14 eV\phi_0 = 2.14 \text{ eV}. When light of frequency ν=6.0×1014 Hz\nu = 6.0 \times 10^{14} \text{ Hz} is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons in eV\text{eV}? (Given h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J})

Solution:

  1. First, calculate the energy of the incident photon in Joules: E=hν=(6.63×10−34 J s)×(6.0×1014 Hz)=39.78×10−20 JE = h\nu = (6.63 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ Hz}) = 39.78 \times 10^{-20} \text{ J}.

  2. Convert this energy into eV\text{eV}: E=39.78×10−201.6×10−19 eV≈2.48 eVE = \frac{39.78 \times 10^{-20}}{1.6 \times 10^{-19}} \text{ eV} \approx 2.48 \text{ eV}.

  3. Use Einstein's photoelectric equation to find KmaxK_{max}: Kmax=E−ϕ0K_{max} = E - \phi_0.

  4. Perform the subtraction: Kmax=2.48 eV−2.14 eVK_{max} = 2.48 \text{ eV} - 2.14 \text{ eV}.

2.48−2.140.34\begin{array}{r} 2.48 \\ - 2.14 \\ \hline 0.34 \end{array}

Kmax=0.34 eVK_{max} = 0.34 \text{ eV}.

Explanation:

According to Einstein's equation, the energy of the incident photon is used in two ways: overcoming the work function (ϕ0\phi_0) and providing maximum kinetic energy (KmaxK_{max}) to the photoelectron. By subtracting the work function from the total photon energy, we find the excess energy available as kinetic energy.

Problem 2:

In a photoelectric effect experiment, the stopping potential for a certain metal is 1.5 V1.5 \text{ V}. What is the maximum kinetic energy of the photoelectrons emitted?

Solution:

The relationship between stopping potential V0V_0 and maximum kinetic energy KmaxK_{max} is given by: Kmax=eV0K_{max} = eV_0 Given V0=1.5 VV_0 = 1.5 \text{ V}, the kinetic energy is: Kmax=e×1.5 V=1.5 eVK_{max} = e \times 1.5 \text{ V} = 1.5 \text{ eV}. To express this in Joules: Kmax=1.5×1.6×10−19 J=2.4×10−19 JK_{max} = 1.5 \times 1.6 \times 10^{-19} \text{ J} = 2.4 \times 10^{-19} \text{ J}.

Explanation:

Stopping potential represents the work done by the electric field to stop the fastest moving electron. Thus, eV0eV_0 is numerically equal to the maximum kinetic energy of the emitted electrons.