NEB Class 12

Physics — Unit 18 / Chapter 20: ELECTRONS

NEB Grade XII • Modern Physics

CHAPTER 20: ELECTRONS

Comprehensive study of subatomic charge carrier discovery, quantization of charge, parabolic & circular trajectories in electric and magnetic fields, and specific charge measurement ($e/m$).

Elementary Charge $e$ $1.602 \times 10^{-19}\,\mathrm{C}$
Electron Mass $m_e$ $9.109 \times 10^{-31}\text{ kg}$
Specific Charge $e/m$ $1.7588 \times 10^{11}\mathrm{C\,kg^{-1}}$
Viscosity of Air $\eta$ $1.81 \times 10^{-5}\mathrm{N\,s\,m^{-2}}$
21.1 Quantum Nature of Radiation & Planck's Hypothesis
21.2 Properties of Photons (Energy, Mass, Momentum, Flux)
21.3 Work Function & Photoelectric Effect Laws
21.4 Einstein's Photoelectric Equation & Derivation
21.5 Millikan's Verification Experiment & Planck's Constant
21.6-7 NEB Solved Numericals & Question Bank
Section 21.1

Quantum Nature of Radiation

1. Failure of Classical Wave Theory

In the 19th century, James Clerk Maxwell’s electromagnetic wave theory successfully explained optical phenomena like interference, diffraction, and polarization. However, classical theory failed fundamentally when applied to Blackbody Radiation Distribution at atomic scales:

  • Wien's Distribution Law: Derived assuming thermal equilibrium of radiation. It successfully explained blackbody radiation at short wavelengths (high frequencies), but failed significantly at longer wavelengths. $$\text{Wien's Formula: } E_\lambda d\lambda = \frac{c_1}{\lambda^5} e^{-c_2 / (\lambda T)} d\lambda$$
  • Rayleigh-Jeans Law: Derived using classical equipartition of energy ($k_B T$ per mode). It matched experimental data at long wavelengths, but predicted that energy density approaches infinity as wavelength approaches zero ($\lambda \to 0$). This catastrophic breakdown at ultraviolet frequencies is famously called the Ultraviolet Catastrophe. $$\text{Rayleigh-Jeans Formula: } E_\lambda d\lambda = \frac{8 \pi k_B T}{\lambda^4} d\lambda$$
Figure 21.1: Blackbody Radiation Energy Distribution Spectrum (Comparison of Theories)
Wavelength $\lambda$ (nm) Energy Density $E_\lambda$ Rayleigh-Jeans (Classical Breakdown) Wien's Law Planck's Experimental Curve UV Region

2. Max Planck's Quantum Hypothesis (1900)

To resolve the breakdown of classical mechanics, German physicist Max Planck proposed a revolutionary hypothesis:

"Emission and absorption of electromagnetic radiation by the atomic oscillators in a blackbody do not occur continuously, but in discrete, localized packets of energy called quanta."

The energy $E$ contained within a single quantum of electromagnetic radiation of frequency $\nu$ is directly proportional to its frequency:

$$E = h \nu = \frac{h c}{\lambda}$$

Where total energy emitted or absorbed by an oscillator is quantized as integral multiples:

$$E_n = n h \nu = n \frac{h c}{\lambda} \quad (n = 1, 2, 3, \dots)$$
Physical Quantity Symbol Standard Value (SI Unit) Electron-Volt Equivalent
Planck's Constant $h$ $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ $4.136 \times 10^{-15} \text{ eV}\cdot\text{s}$
Speed of Light in Vacuum $c$ $3.00 \times 10^{8} \text{ m/s}$ —
Product $h \cdot c$ $hc$ $1.988 \times 10^{-25} \text{ J}\cdot\text{m}$ $1240 \text{ eV}\cdot\text{nm}$
Elementary Charge $e$ $1.602 \times 10^{-19} \text{ C}$ $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$
Section 21.2

Properties of Photons

In 1905, Albert Einstein extended Planck's quantum concept to state that light itself propagates through space as discrete localized energy packets called Photons. The key fundamental properties of photons required for NEB examinations are detailed below:

1 Energy Quantization

Every photon carries a discrete quantum of energy proportional to frequency:

$$E = h\nu = \frac{hc}{\lambda}$$

2 Invariant Speed

In vacuum, all photons travel at the speed of light $c = 3 \times 10^8 \text{ m/s}$, regardless of the motion of the source or observer.

3 Rest Mass & Relativistic Mass

The rest mass of a photon is strictly zero ($m_0 = 0$). According to Einstein's mass-energy equivalence $E = m c^2$, its dynamic relativistic mass is:

$$m = \frac{E}{c^2} = \frac{h\nu}{c^2} = \frac{h}{c\lambda}$$

4 Linear Momentum

Although rest mass is zero, a moving photon possesses finite linear momentum $p$:

$$p = m c = \left(\frac{h\nu}{c^2}\right) c = \frac{h\nu}{c} = \frac{h}{\lambda}$$

5 Electrical Neutrality

Photons are electrically neutral. They are not deflected by electric field $\vec{E}$ or magnetic field $\vec{B}$.

6 Particle-Particle Collisions

In photon-electron collisions (e.g., Compton effect), total energy and total momentum are conserved. However, total number of photons may not be conserved (photons can be absorbed or created).

3. Photon Intensity and Photon Flux Derivation

Consider a monochromatic light source of power $P$ emitting photons of frequency $\nu$ uniformly.

  1. Number of Photons Emitted per Second ($N$): $$P = \frac{\text{Total Energy}}{\text{time}} = \frac{N \cdot E}{1\text{ s}} = N h \nu \implies N = \frac{P}{h\nu} = \frac{P \lambda}{hc}$$
  2. Radiation Intensity ($I$): Defined as light power per unit cross-sectional area $A$: $$I = \frac{P}{A} = \left(\frac{N}{A}\right) h \nu = n_s \cdot h \nu$$ Where $n_s = \frac{N}{A}$ is the photon flux density (number of photons falling per unit area per second).
Section 21.3

Work Function and Photoelectric Effect

1. Key Terminology Definitions

A. Photoelectric Effect:
The phenomenon of ejection of electrons from a clean metal surface when electromagnetic radiation (light) of suitable frequency strikes the surface. Emitted electrons are called photoelectrons and the resulting current is called photocurrent.
B. Work Function ($\phi$ or $W_0$):
The minimum energy required to liberate a bound electron from the surface of a given metal against atomic attractive forces with zero remaining kinetic energy. $$\phi = h \nu_0 = \frac{h c}{\lambda_0}$$
C. Threshold Frequency ($\nu_0$):
The minimum characteristic frequency of incident radiation below which no photoelectric emission occurs, no matter how intense the incident beam is.
D. Threshold Wavelength ($\lambda_0$):
The maximum wavelength of incident light capable of ejecting photoelectrons from a given metal surface ($\lambda_0 = c / \nu_0$).
E. Stopping Potential / Cut-off Voltage ($V_0$):
The minimum retarding (negative) potential applied to the collector anode plate relative to the cathode emitter plate that is just sufficient to stop the fastest-moving photoelectrons from reaching the collector, reducing the photocurrent strictly to zero ($I = 0$). $$e V_0 = K_{\max} = \frac{1}{2} m v_{\max}^2$$

Table 21.1: Work Functions of Representative Metal Elements

Metal Surface Work Function $\phi$ (eV) Work Function $\phi$ ($10^{-19}\text{ J}$) Threshold $\lambda_0$ (nm) Light Sensitivity Range
Cesium (Cs) 2.14 eV 3.43 J 579 nm Visible (Yellow-Green)
Potassium (K) 2.30 eV 3.68 J 539 nm Visible (Green)
Sodium (Na) 2.36 eV 3.78 J 525 nm Visible (Green)
Zinc (Zn) 4.31 eV 6.90 J 288 nm Ultraviolet (UV)
Platinum (Pt) 5.65 eV 9.05 J 219 nm Deep UV

2. Experimental Apparatus Setup

The standard laboratory apparatus for studying photoelectric phenomena consists of an evacuated glass/quartz tube containing two metallic electrodes:

Figure 21.2: Standard Experimental Setup for Photoelectric Effect Investigation
Monochromatic Light Source Evacuated Quartz Glass Envelope Cathode (C) Anode (A) Ejected Photoelectrons ($e^-$) Reversing Key (Commutator) V μA Variable DC Voltage Source

3. Laws of Photoelectric Emission (Experimental Observations)

By systematically varying parameters in the setup above, Lenard and Millikan established four fundamental laws:

Law I: Effect of Intensity on Photocurrent

At a fixed frequency above threshold ($\nu > \nu_0$) and constant accelerating potential, the photoelectric current is directly proportional to the intensity of incident light.

Law II: Independence of Maximum Kinetic Energy from Intensity

For a given frequency and target material, the maximum kinetic energy $K_{\max}$ (and stopping potential $V_0$) of photoelectrons is completely independent of light intensity.

Law III: Frequency Dependence & Threshold Frequency

Maximum kinetic energy $K_{\max}$ increases linearly with incident frequency $\nu$. If $\nu < \nu_0$, no photoelectrons are ejected, regardless of beam intensity or duration.

Law IV: Instantaneous Nature

Photoelectric emission is an instantaneous process. The time lag between the striking of light photons and electron emission is less than $10^{-9}\text{ seconds}$ ($1\text{ ns}$).

Figure 21.3: Fundamental Characteristic Curves of Photoelectric Effect
Anode Potential ($V$) Photocurrent ($I$) $-V_0$ Intensity $I_3$ (High) Intensity $I_2$ Intensity $I_1$ (Low)

(a) Photocurrent vs Potential (Constant frequency $\nu$)

$V$ $I$ $-V_{03}$ $-V_{02}$ $-V_{01}$ Saturation Current (Same Intensity)

(b) Photocurrent vs Potential (Frequencies $\nu_3 > \nu_2 > \nu_1$)

Section 21.4

Einstein's Photoelectric Equation

1. Theoretical Derivation

In 1905, Albert Einstein successfully explained the photoelectric effect by applying Planck's quantum hypothesis. He postulated that photoelectric interaction is a one-to-one collision between a single incident light photon of energy $h\nu$ and a bound electron in the metal.

When a photon strikes a surface electron, its energy $h\nu$ is completely transferred to the electron. This energy is expended in two parts:

  1. Work Function ($\phi$): The minimum energy used to pull the electron out of the metal surface against atomic binding forces.
  2. Maximum Kinetic Energy ($K_{\max}$): The remaining energy imparted to the emitted electron as kinetic energy: $$K_{\max} = \frac{1}{2} m v_{\max}^2$$

By Conservation of Energy:

$$h \nu = \phi + K_{\max}$$

Since $\phi = h \nu_0$ (where $\nu_0$ is threshold frequency):

$$h \nu = h \nu_0 + \frac{1}{2} m v_{\max}^2$$ $$\implies K_{\max} = \frac{1}{2} m v_{\max}^2 = h(\nu - \nu_0) = h c \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$$

Relating $K_{\max}$ to the stopping potential $V_0$:

$$e V_0 = h\nu - \phi = h(\nu - \nu_0)$$

2. Explanation of Photoelectric Laws using Einstein's Equation

Explanation of Threshold Frequency ($\nu_0$): From $K_{\max} = h(\nu - \nu_0)$, if incident frequency $\nu < \nu_0$, kinetic energy $K_{\max}$ becomes negative, which is physically impossible. Hence, no emission occurs below $\nu_0$.
Explanation of $K_{\max}$ Linear Dependence: $K_{\max}$ depends linearly on frequency $\nu$ via slope $h$, independent of intensity.
Explanation of Intensity Effect: Increasing light intensity increases the number of photons per second, yielding more ejected photoelectrons per second (higher photocurrent), but individual photon energy $h\nu$ remains unchanged.
Explanation of Instantaneous Emission: Energy transfer occurs in a single quantum collision; energy is delivered instantly without waiting for wave absorption.
Section 21.5

Millikan's Verification & Planck's Constant Determination

1. Experimental Principle (1916)

American physicist Robert A. Millikan performed precise experiments using clean alkali metals (Lithium, Sodium, Potassium) inside a ultra-high vacuum cylinder. His objective was to test Einstein's photoelectric equation and measure Planck's constant $h$.

Figure 21.4: Millikan's Vacuum Chamber Setup for Photoelectric Verification
Ultra-High Vacuum Glass Cylinder Rotating Wheel W Knife Scraper K (Cleaning Surface) Monochromatic UV Light Faraday Cup Collector (C) Electrometer

2. Mathematical Analysis & Graphical Method

Rearranging Einstein's equation in terms of stopping potential $V_0$:

$$e V_0 = h \nu - \phi \implies V_0 = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e}$$

This matches the equation of a straight line $y = m x + c$:

  • Variable $y$: Stopping potential $V_0$ (Volts)
  • Variable $x$: Incident light frequency $\nu$ ($\text{Hz}$)
  • Slope ($m$): $m = \frac{h}{e}$ (Universal constant for all target metals!)
  • x-intercept ($\nu$-axis): Threshold frequency $\nu_0$ (where $V_0 = 0$)
  • y-intercept ($V_0$-axis): Retarding potential intercept $-\frac{\phi}{e}$
Figure 21.5: Millikan's Stopping Potential ($V_0$) vs Frequency ($\nu$) Verification Graph
Frequency $\nu$ ($10^{14}\text{ Hz}$) Stopping Potential $V_0$ (V) $\nu_0$ (Sodium) $-\frac{\phi}{e}$ $\nu_0'$ (Zinc) $\Delta \nu$ $\Delta V_0$ $\text{Slope } m = \frac{\Delta V_0}{\Delta \nu} = \frac{h}{e}$

3. Extraction of Planck's Constant ($h$)

By calculating the experimental slope $m$ from the $V_0$ vs $\nu$ straight line and multiplying by the known charge of an electron $e = 1.602 \times 10^{-19}\text{ C}$:

$$h = e \times \text{Slope} = e \times \left( \frac{\Delta V_0}{\Delta \nu} \right)$$

Millikan obtained $h = 6.57 \times 10^{-34} \text{ J s}$, which agreed with Planck's radiation constant within 0.5% error, conclusively proving Einstein's photon hypothesis.

Interactive Laboratory Suite

Physics Simulations & Calculators

Explore photoelectric emission physics, plot experimental Millikan graphs, and calculate photon parameters in real time.

⚡ Simulation 1: Photoelectric Effect Virtual Tube Experiment

Live Electron Trajectory
Near UV
Controls Photon Emission Rate
Negative = Retarding Potential
Photon Energy ($E$) 3.10 eV
Work Function ($\phi$) 2.36 eV
Max $K_{\max}$ / $V_0$ 0.74 eV / 0.74 V
Photocurrent ($I$) 12.4 μA

📈 Simulation 2: Millikan $V_0$ vs Frequency Graph & $h$ Extractor

Experimental Data Fitting

Recorded Data Points ($\nu, V_0$)

🧮 Simulation 3: Complete Photon Physics Parameter Calculator

Photon Frequency ($\nu$) 6.00 × 10¹⁴ Hz
Energy ($E$) 2.48 eV
Momentum ($p$) 1.33 × 10⁻²⁷ kg m/s
Photon Flux ($N/t$) 2.52 × 10²⁰ /sec
Section 21.6

Step-by-Step Solved Numericals (NEB Standard)

Numerical Problem 21.1 NEB 2078 Model Question

Calculate the energy (in Joules and eV) and linear momentum of a single light photon having a wavelength of $\lambda = 400\text{ nm}$.

Step-by-Step Solution:

Given Data:

Wavelength $\lambda = 400\text{ nm} = 400 \times 10^{-9}\text{ m}$

Planck's constant $h = 6.626 \times 10^{-34}\text{ J s}$, Speed of light $c = 3.0 \times 10^8\text{ m/s}$

1. Photon Energy ($E$):

$$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{400 \times 10^{-9}} = 4.9695 \times 10^{-19}\text{ Joules}$$

In electron-volts ($\text{eV}$):

$$E_{\text{eV}} = \frac{4.9695 \times 10^{-19}\text{ J}}{1.602 \times 10^{-19}\text{ J/eV}} = 3.102\text{ eV}$$

2. Photon Momentum ($p$):

$$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}\text{ J s}}{400 \times 10^{-9}\text{ m}} = 1.6565 \times 10^{-27}\text{ kg m/s}$$

✅ Final Answer: Energy $= 4.97 \times 10^{-19}\text{ J} = 3.10\text{ eV}$; Momentum $= 1.66 \times 10^{-27}\text{ kg m/s}$.

Numerical Problem 21.2 NEB 2076 Board Exam

Light of wavelength $300\text{ nm}$ falls on a sodium plate having a work function of $\phi = 2.36\text{ eV}$. Find: (a) Threshold wavelength of sodium, (b) Maximum kinetic energy of photoelectrons (in eV), and (c) Maximum emission velocity of photoelectrons.

Step-by-Step Solution:

Given Data:

Incident Wavelength $\lambda = 300\text{ nm} = 300 \times 10^{-9}\text{ m}$

Work Function $\phi = 2.36\text{ eV} = 2.36 \times 1.602 \times 10^{-19}\text{ J} = 3.7807 \times 10^{-19}\text{ J}$

Electron mass $m = 9.109 \times 10^{-31}\text{ kg}$

(a) Threshold Wavelength ($\lambda_0$):

$$\phi = \frac{hc}{\lambda_0} \implies \lambda_0 = \frac{hc}{\phi} = \frac{1.9878 \times 10^{-25}\text{ J m}}{3.7807 \times 10^{-19}\text{ J}} = 5.257 \times 10^{-7}\text{ m} = 525.7\text{ nm}$$

(b) Maximum Kinetic Energy ($K_{\max}$):

Incident Photon Energy $E = \frac{hc}{\lambda} = \frac{1240\text{ eV nm}}{300\text{ nm}} = 4.133\text{ eV}$

$$K_{\max} = E - \phi = 4.133\text{ eV} - 2.360\text{ eV} = 1.773\text{ eV}$$

In Joules: $K_{\max} = 1.773 \times 1.602 \times 10^{-19}\text{ J} = 2.840 \times 10^{-19}\text{ J}$

(c) Maximum Emission Velocity ($v_{\max}$):

$$K_{\max} = \frac{1}{2} m v_{\max}^2 \implies v_{\max} = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 \times 2.840 \times 10^{-19}}{9.109 \times 10^{-31}}} = 7.886 \times 10^{5}\text{ m/s}$$

✅ Final Answer: $\lambda_0 = 525.7\text{ nm}$, $K_{\max} = 1.77\text{ eV}$, $v_{\max} = 7.89 \times 10^5\text{ m/s}$.

Numerical Problem 21.3 NEB Standard High Weightage

Ultraviolet light of frequency $\nu = 1.2 \times 10^{15}\text{ Hz}$ falls on a metal surface. The stopping potential required to stop photoelectron current is $V_0 = 2.80\text{ V}$. Calculate the work function of the metal in eV and determine its threshold frequency.

Step-by-Step Solution:

Given Data:

Frequency $\nu = 1.2 \times 10^{15}\text{ Hz}$

Stopping Potential $V_0 = 2.80\text{ V} \implies K_{\max} = e V_0 = 2.80\text{ eV}$

1. Calculate Incident Photon Energy ($E$):

$$E = h\nu = (4.136 \times 10^{-15}\text{ eV s}) \times (1.2 \times 10^{15}\text{ Hz}) = 4.963\text{ eV}$$

2. Calculate Work Function ($\phi$):

$$e V_0 = E - \phi \implies \phi = E - e V_0 = 4.963\text{ eV} - 2.800\text{ eV} = 2.163\text{ eV}$$

3. Calculate Threshold Frequency ($\nu_0$):

$$\nu_0 = \frac{\phi}{h} = \frac{2.163\text{ eV}}{4.136 \times 10^{-15}\text{ eV s}} = 5.23 \times 10^{14}\text{ Hz}$$

✅ Final Answer: Work Function $\phi = 2.16\text{ eV}$; Threshold Frequency $\nu_0 = 5.23 \times 10^{14}\text{ Hz}$.

Numerical Problem 21.4 Millikan Graph Analysis

In a Millikan-type photoelectric experiment, stopping potentials measured at two different monochromatic frequencies are: $V_{01} = 1.85\text{ V}$ at $\nu_1 = 1.0 \times 10^{15}\text{ Hz}$, and $V_{02} = 0.61\text{ V}$ at $\nu_2 = 7.0 \times 10^{14}\text{ Hz}$. Calculate the experimental value of Planck's constant $h$ and work function $\phi$.

Step-by-Step Solution:

Using $e V_0 = h \nu - \phi$ for both frequencies:

1) $e V_{01} = h \nu_1 - \phi$

2) $e V_{02} = h \nu_2 - \phi$

Subtracting equation (2) from equation (1):

$$e (V_{01} - V_{02}) = h (\nu_1 - \nu_2) \implies h = e \times \frac{V_{01} - V_{02}}{\nu_1 - \nu_2}$$

$$\Delta V_0 = 1.85 - 0.61 = 1.24\text{ V}$$

$$\Delta \nu = (10.0 - 7.0) \times 10^{14} = 3.0 \times 10^{14}\text{ Hz}$$

$$h = (1.602 \times 10^{-19}\text{ C}) \times \frac{1.24\text{ V}}{3.0 \times 10^{14}\text{ Hz}} = 6.6216 \times 10^{-34}\text{ J s}$$

Now finding $\phi$ from eq (1):

$$\phi = h\nu_1 - e V_{01} = (4.133 \times 10^{-15} \times 10^{15}) - 1.85\text{ eV} = 4.133 - 1.85 = 2.283\text{ eV}$$

✅ Final Answer: $h = 6.62 \times 10^{-34}\text{ J s}$; Work Function $\phi = 2.28\text{ eV}$.

Numerical Problem 21.5 Photon Flux & Rate

A $100\text{ W}$ sodium lamp radiates light uniformly in all directions at wavelength $\lambda = 589\text{ nm}$. Assuming 5% of electric power is converted into light, calculate: (a) Total rate of photon emission per second, and (b) Photon flux hitting a target plate of area $1\text{ cm}^2$ placed $2\text{ meters}$ away.

Step-by-Step Solution:

Light Output Power $P = 5\% \times 100\text{ W} = 5.0\text{ Watts} = 5.0\text{ J/s}$

(a) Single Photon Energy ($E$):

$$E = \frac{hc}{\lambda} = \frac{1.988 \times 10^{-25}\text{ J m}}{589 \times 10^{-9}\text{ m}} = 3.375 \times 10^{-19}\text{ Joules}$$

Number of photons emitted per second ($N$):

$$N = \frac{P}{E} = \frac{5.0\text{ J/s}}{3.375 \times 10^{-19}\text{ J}} = 1.481 \times 10^{19}\text{ photons/second}$$

(b) Photon Flux at Distance $r = 2\text{ m}$:

Area of sphere at $r=2\text{ m}$: $A_{\text{sphere}} = 4 \pi r^2 = 4 \pi (2)^2 = 50.265\text{ m}^2$

Target area $A_{\text{target}} = 1\text{ cm}^2 = 10^{-4}\text{ m}^2$

Photons hitting target per second:

$$n_{\text{target}} = N \times \frac{A_{\text{target}}}{A_{\text{sphere}}} = (1.481 \times 10^{19}) \times \frac{10^{-4}}{50.265} = 2.946 \times 10^{13}\text{ photons/s}$$

✅ Final Answer: Emission Rate $= 1.48 \times 10^{19}\text{ s}^{-1}$; Target Flux $= 2.95 \times 10^{13}\text{ photons/s}$.

Section 21.7

NEB Practice Question Bank

Q1. Why is wave theory of light unable to explain the photoelectric effect?

Answer: Wave theory predicts that kinetic energy of emitted photoelectrons should increase with light intensity and that emission should require time for wave energy accumulation. Experimentally, $K_{\max}$ is independent of intensity, and emission is instantaneous ($<10^{-9}\text{ s}$). Furthermore, wave theory cannot account for threshold frequency.

Q2. Define stopping potential. Does it depend on the intensity of light?

Answer: Stopping potential $V_0$ is the minimum retarding potential required to reduce photocurrent to zero. No, $V_0$ is completely independent of light intensity; it depends solely on incident frequency $\nu$ and work function $\phi$.

Q3. What is the rest mass and effective dynamic mass of a photon?

Answer: The rest mass of a photon is zero ($m_0 = 0$). Its dynamic relativistic mass when moving at speed $c$ is $m = \frac{E}{c^2} = \frac{h\nu}{c^2} = \frac{h}{c\lambda}$.

Q4. Alkali metals like Cesium and Potassium are preferred in photocells. Why?

Answer: Alkali metals have very low work functions ($\phi \approx 2.0 - 2.3\text{ eV}$), enabling them to emit photoelectrons when illuminated by visible light, whereas metals like Zinc require higher-energy UV radiation.

Q5. What is the physical significance of the slope of $V_0$ versus frequency ($\nu$) graph?

Answer: The slope of the $V_0$ vs $\nu$ graph represents the universal ratio $\frac{h}{e}$. Multiplying this slope by electron charge $e$ yields Planck's constant $h$.