Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below:

Related Formula

where, K$K_{\max}$ = maximum kinetic energy of photoelectrons Vₛ$V_s$ = stopping potential e$e$ = fundamental electronic charge

Core Logic

By definition, the stopping potential Vₛ$V_s$ is the negative potential applied to stop the most energetic photoelectrons from reaching the collector electrode.

From Einstein's photoelectric equation:

Thus, the stopping potential is exactly (1)/(e)$\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons. It does not depend on the intensity of light.

Pattern Recognition

Remember the primary features of the photoelectric effect:

• Stopping potential depends linearly on frequency, and inversely on wavelength.
• Intensity changes current, but has zero effect on stopping potential.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Core Logic

Let E$\mathrm{E}$ represent the identical energy value shared by both particles:

Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength:

Step 1: Final Conclusion

The calculated ratio maps to option (3).

Pattern Recognition

Combine the core formulas: λmatter = h 2mE$\lambda_{\text{matter}} = \frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$ and λlight = hcmathrmE$\lambda_{\text{light}} = \frac{\mathrm{hc}}{mathrm{E}}$. Dividing them smoothly yields the standard non-relativistic scaling ratio.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Einstein's Photoelectric Equation:

For emission to occur, the photon energy must exceed the work function:

Useful shortcut: hc ≈ 1240~eV· nm$hc \approx 1240\mathrm{~eV\cdot nm}$.

Core Logic

Let's find the threshold wavelength λthreshold$\lambda_{\text{threshold}}$ for a work function of φ = 3~eV$\phi = 3\mathrm{~eV}$:

To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold:

Step 1: Comparing Visible Spectrum Colors

Let's look at the standard approximate wavelength ranges for visible light:

• Red: 620 - 750~nm$620 - 750\mathrm{~nm}$ (Energy ≈ 1.65 - 2.0~eV$\approx 1.65 - 2.0\mathrm{~eV}$)
• Yellow: 570 - 590~nm$570 - 590\mathrm{~nm}$ (Energy ≈ 2.1 - 2.2~eV$\approx 2.1 - 2.2\mathrm{~eV}$)
• Green: 495 - 570~nm$495 - 570\mathrm{~nm}$ (Energy ≈ 2.2 - 2.5~eV$\approx 2.2 - 2.5\mathrm{~eV}$)
• Blue: 450 - 495~nm$450 - 495\mathrm{~nm}$ (At lower end, approaching violet down to 380~nm$380\mathrm{~nm}$; Energy ≈ 2.5 - 3.3~eV$\approx 2.5 - 3.3\mathrm{~eV}$)

Only Blue light contains wavelengths extending below 413.3~nm$413.3\mathrm{~nm}$ (high enough photon energy to surpass 3~eV$3\mathrm{~eV}$). Therefore, blue light is required to cause photoelectric emission from this metal.

Pattern Recognition

Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically > 2.8~eV$> 2.8\mathrm{~eV}$), while Red/Yellow photons have much lower energy (< 2.2~eV$< 2.2\mathrm{~eV}$). For a higher work function like 3~eV$3\mathrm{~eV}$, only highly energetic blue/violet light can succeed.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Force due to a completely reflecting beam:

Change in momentum:

Work-Energy Theorem or Conservation of Mechanical Energy for small deflections:

For small angles (θ)/(2) ≈ (θ)/(2)$\sin\frac{\theta}{2} \approx \frac{\theta}{2}$.

Core Logic

Assuming perfect normal reflection from the mirror surface, the pulse imparts a momentum impulse of (2E)/(c)$\frac{2E}{c}$ to the mass. This provides an initial velocity v$v$ to the mirror. The mirror then swings up to a maximum angle θ$\theta$ where kinetic energy converts entirely to gravitational potential energy.

Step 1: Calculate Initial Velocity

From momentum change:

Step 2: Relate to Angular Deflection

Using conservation of mechanical energy:

Since θ$\theta$ is very small, (θ)/(2) ≈ (θ)/(2)$\sin\frac{\theta}{2} \approx \frac{\theta}{2}$:

Step 3: Solve for Theta

Substitute v = (2E)/(mc)$v = \frac{2E}{mc}$ into the expression:

Pattern Recognition

This is a standard ballistic pendulum problem where the impulse is delivered by radiation pressure. Perfect reflection means momentum transfer is double the incident momentum (2· (E)/(c))$\left(2\cdot \frac{E}{c}\right)$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Einstein's photoelectric equation:

eVₛ = h u - φ$$

eV_s = h u - \phi$$

where:

• Vₛ$V_s$ = stopping potential

u$

u$ = frequency of light

• φ$\phi$ = work function
• Intensity formula: I = (n h

u)/(A · t)$I = \frac{n h

u}{A \cdot t}$ (where n$n$ is rate of photons).

Core Logic

• Assertion Analysis: Stopping potential Vₛ$V_s$ depends strictly linearly on frequency

u$

u$ and work function φ$\phi$. It is completely independent of the beam intensity. Therefore, Assertion A is false.

• Reason Analysis: Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given

u > u₀$

u > u_0$. Thus, Reason R is true.

Pattern Recognition

Stopping Potential rightarrow$\leftrightarrow$ Frequency/Energy characteristic. Photo-current / Emission Rate rightarrow$\leftrightarrow$ Photon Intensity/Flux counts.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter