Solution Q13: Quantum Yield
Using the data from the graph (Saturation current $6.0 \mu A$) and calculating the incident photon flux using $P = \frac{1}{2}\epsilon_0 c E^2 A \cos 30^\circ$ and $\omega$. The calculated efficiency matches option (b).
Correct Option: (b) 0.038
Solution Q14: Amplitude Change
Saturation current is proportional to Intensity ($I \propto E_0^2$).
$$ i_{new} = i_{old} \times \left( \frac{E_{new}}{E_{old}} \right)^2 $$ $$ i_{new} = 6.0 \mu A \times \left( \frac{25}{15} \right)^2 = 6.0 \times \frac{25}{9} = \frac{150}{9} \approx 16.67 \mu A $$Correct Option: (c) 16.67 $\mu A$
Solution Q15: Complex Wave Analysis
Key Concept 1 (Stopping Potential): The stopping potential is determined by the maximum kinetic energy of the photoelectrons, which depends on the highest frequency component present in the incident light ($K_{max} = h\nu_{max} – \phi$).
Key Concept 2 (Saturation Current): The saturation current is proportional to the rate of emission of photoelectrons, which is proportional to the flux of incident photons ($N$). $$ N = \frac{\text{Intensity}}{h\nu} $$ For a wave with multiple frequency components, we must calculate the photon flux for each component separately.
Key Concept 2 (Saturation Current): The saturation current is proportional to the rate of emission of photoelectrons, which is proportional to the flux of incident photons ($N$). $$ N = \frac{\text{Intensity}}{h\nu} $$ For a wave with multiple frequency components, we must calculate the photon flux for each component separately.
Step 1: Analyze the Electric Field Equation
Given: $E = E_0 [1 + \cos(\omega t + \phi_1)] \cos(\omega t + \phi_2)$
Expand using the identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$: $$ E = E_0 \cos(\omega t + \phi_2) + E_0 \cos(\omega t + \phi_1)\cos(\omega t + \phi_2) $$ $$ E = E_0 \cos(\omega t + \phi_2) + \frac{E_0}{2} \cos(2\omega t + \phi_1 + \phi_2) + \frac{E_0}{2} \cos(\phi_1 – \phi_2) $$ This wave has three components:
Given: $E = E_0 [1 + \cos(\omega t + \phi_1)] \cos(\omega t + \phi_2)$
Expand using the identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$: $$ E = E_0 \cos(\omega t + \phi_2) + E_0 \cos(\omega t + \phi_1)\cos(\omega t + \phi_2) $$ $$ E = E_0 \cos(\omega t + \phi_2) + \frac{E_0}{2} \cos(2\omega t + \phi_1 + \phi_2) + \frac{E_0}{2} \cos(\phi_1 – \phi_2) $$ This wave has three components:
- Fundamental ($\omega$): Amplitude $A_1 = E_0$. Energy per photon $E_{ph} = \hbar\omega$.
- Second Harmonic ($2\omega$): Amplitude $A_2 = \frac{E_0}{2}$. Energy per photon $E’_{ph} = 2\hbar\omega$.
- DC Component: Amplitude $\frac{E_0}{2}\cos(\dots)$. This is a static field and does not contribute to the photoelectric effect (frequency = 0).
Step 2: Calculate New Stopping Potential
The stopping potential is determined by the highest frequency component, which is $2\omega$. From the graph (Original Wave $\omega$): $$ eV_{stop1} = 2.0 \text{ eV} \implies \hbar\omega – \phi = 2.0 \text{ eV} $$ For the New Wave ($2\omega$): $$ eV_{stop2} = (2\hbar\omega) – \phi $$ Rewrite as: $$ eV_{stop2} = 2(\hbar\omega – \phi) + \phi = 2(2.0) + \phi = 4.0 + \phi $$ We need to find $\phi$. Using $\omega = 9.5 \times 10^{14}$ rad/s: $$ \hbar\omega \approx \frac{6.63 \times 10^{-34} \times 9.5 \times 10^{14}}{1.6 \times 10^{-19} \times 2\pi} \approx 6.27 \text{ eV} $$ $$ \phi = \hbar\omega – 2.0 = 6.27 – 2.0 = 4.27 \text{ eV} $$ Substitute $\phi$ back: $$ V_{stop2} = 2(6.27) – 4.27 = 12.54 – 4.27 = 8.27 \text{ V} $$
The stopping potential is determined by the highest frequency component, which is $2\omega$. From the graph (Original Wave $\omega$): $$ eV_{stop1} = 2.0 \text{ eV} \implies \hbar\omega – \phi = 2.0 \text{ eV} $$ For the New Wave ($2\omega$): $$ eV_{stop2} = (2\hbar\omega) – \phi $$ Rewrite as: $$ eV_{stop2} = 2(\hbar\omega – \phi) + \phi = 2(2.0) + \phi = 4.0 + \phi $$ We need to find $\phi$. Using $\omega = 9.5 \times 10^{14}$ rad/s: $$ \hbar\omega \approx \frac{6.63 \times 10^{-34} \times 9.5 \times 10^{14}}{1.6 \times 10^{-19} \times 2\pi} \approx 6.27 \text{ eV} $$ $$ \phi = \hbar\omega – 2.0 = 6.27 – 2.0 = 4.27 \text{ eV} $$ Substitute $\phi$ back: $$ V_{stop2} = 2(6.27) – 4.27 = 12.54 – 4.27 = 8.27 \text{ V} $$
Step 3: Calculate New Saturation Current
Current depends on total photon flux $N_{total}$. Intensity $I$ is proportional to (Amplitude)$^2$.
Current depends on total photon flux $N_{total}$. Intensity $I$ is proportional to (Amplitude)$^2$.
- Original Flux ($N_0$): Only frequency $\omega$, Amplitude $E_0$. $$ I_0 \propto E_0^2 \quad \Rightarrow \quad N_0 \propto \frac{E_0^2}{\hbar\omega} $$
- New Flux ($N_{new}$): Sum of fluxes from $\omega$ and $2\omega$ components.
1. Component $\omega$ (Amp $E_0$): $$ I_1 \propto E_0^2 \implies N_1 \propto \frac{E_0^2}{\hbar\omega} $$ 2. Component $2\omega$ (Amp $E_0/2$): $$ I_2 \propto \left(\frac{E_0}{2}\right)^2 = \frac{E_0^2}{4} $$ $$ N_2 \propto \frac{I_2}{2\hbar\omega} = \frac{E_0^2/4}{2\hbar\omega} = \frac{1}{8} \frac{E_0^2}{\hbar\omega} $$ Note: We divide by $2\hbar\omega$ because each photon carries twice the energy.
Correct Option: (c) 8.27 V and 6.75 $\mu A$