A photo-emissive substance is illuminated with a radiation of wavelength λᵢ$\lambda_{i}$ so that it releases electrons with de-Broglie wavelength λc$\lambda_{c}$ The longest wavelength of radiation that can emit photoelectron is λ₀$\lambda_{0}$. Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light) [cite: 39, 41]

Related Formula

h u = φ₀ + K = φ₀ + eV₀$$h u = \phi_0 + K_{\text{\max}} = \phi_0 + eV_0$$

V₀ = ((h)/(e)) u - (φ₀)/(e)$$V_0 = \left(\frac{h}{e}\right) u - \frac{\phi_0}{e}$$

Core Logic

Analyzing each statement:

• (A) V₀ v/s

u$V_0 \text{ v/s }

u$ is a straight line equation (y = mx + c$y = mx + c$). True.

• (B) Slope is (h)/(e)$\frac{h}{e}$, not (φ₀)/(h)$\frac{\phi_0}{h}$. False.
• (C) The slope involves Planck's constant h$h$. True.
• (D) To find h$h$ from the slope (m = h/e$m = h/e$), you must multiply the slope by the electronic charge e$e$. Thus, e$e$ is required. False.
• (E) The

u$

u$-intercept occurs when V₀ = 0 uth = (φ₀)/(h)$V_0 = 0 \implies u_{\text{th}} = \frac{\phi_0}{h}$. The y$y$-intercept is -φ₀/e$-\phi_0/e$. Therefore, we can find φ₀$\phi_0$ from the intercepts knowing only e$e$ or by looking at the scale parameters carefully, without explicitly knowing h$h$. True.

Hence, statements (A), (C), and (E) are correct.

Pattern Recognition

Einstein's photoelectric equation yields a straight-line plot for V₀$V_0$ vs u$ u$ where slope is universally (h)/(e)$\frac{h}{e}$ for all metals.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Relativistic total energy equation:

Core Logic

We can verify this equation via dimensional analysis:

• Dimension of energy, [E] = M¹ L² T⁻² [E²] = M² L⁴ T⁻⁴$[E] = M^1 L^2 T^{-2} \implies [E^2] = M^2 L^4 T^{-4}$
• Dimension of momentum times light speed, [pc] = (M¹ L¹ T⁻¹) · (L¹ T⁻¹) = M¹ L² T⁻² [p²c²] = M² L⁴ T⁻⁴$[pc] = (M^1 L^1 T^{-1}) \cdot (L^1 T^{-1}) = M^1 L^2 T^{-2} \implies [p^2c^2] = M^2 L^4 T^{-4}$
• Rest energy square term, [m²c⁴] = M² · (L¹ T⁻¹)⁴ = M² L⁴ T⁻⁴$[m^2c^4] = M^2 \cdot (L^1 T^{-1})^4 = M^2 L^4 T^{-4}$

Since all terms share identical dimensions, this relativistic identity is structurally valid. The matching option is (4).

Pattern Recognition

This is Einstein's famous energy-momentum relation of special relativity, fundamental to high-energy modern physics contexts.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Power of a light source:

where n$n$ is the number of photons emitted per second.

Core Logic

Taking the ratio of power for source 1 and source 2:

Given data:

• (P₁)/(P₂) = 2$\frac{P_1}{P_2} = 2$
• n₁ = 2 × 10¹⁵ photons/s$n_1 = 2 \times 10^{15}\ \mathrm{photons/s}$
• λ₁ = 600 nm$\lambda_1 = 600\ \mathrm{nm}$
• λ₂ = 300 nm$\lambda_2 = 300\ \mathrm{nm}$

Substitute the values:

The question asks for the value multiplying 10¹⁴$10^{14}$, which is 5.

Pattern Recognition

Power relies on both the delivery rate and individual photon packet energy. Lower wavelength means more energetic packets, requiring fewer photons to emit the same power.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

The de Broglie wavelength relation matching a moving particle momentum is given by:

Core Logic

Calculate the electric acceleration force acting component on the charge :

Applying kinematics to find velocity at time t$t$:

Step 1: Calculating Velocity Magnitude and Final Wavelength

Find the magnitude of the updated velocity vector:

Substitute this into the wavelength equation:

Since initial wavelength matches λ₀ = hmv₀$\lambda_{0} = \frac{h}{mv_{0}}$, the expression simplifies to :

Pattern Recognition

The perpendicular field increases the particle's overall velocity and momentum. Since wavelength is inversely proportional to momentum, it must decrease, which rules out options with a plus sign in the numerator.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter