a) A graph between stopping potential ( V o ) with the frequency (f) of incident radiation is shown in figure below. Answer the following questions. [figure in the original paper] i) Write the…

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a) Photoelectric Effect

i) Einstein photoelectric equation in terms of stopping potential

The Einstein photoelectric equation relates the maximum kinetic energy ($K_{\text{max}}$) of emitted photoelectrons to the frequency ($f$) of incident radiation and the work function ($\varphi$) of the metal:

$$K_{\text{max}}=hf-\varphi$$

At the stopping potential ($V_o$), the maximum kinetic energy of the photoelectrons is equal to the work done against the potential barrier:

$$e{V}_o=K_{\text{max}}=hf-\varphi$$

Thus, the equation in terms of stopping potential is:

$$\boxed{e{V}_o=hf-\varphi }$$

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ii) Determination of Planck’s constant ($h$) from the graph

From the graph of $V_o$ vs. $f$, we observe a linear relationship of the form:

$$V_o=\left(\frac{h}{e}\right)f-\frac{\varphi }{e}$$

• Slope ($m$): The slope of the line gives $\frac{h}{e}$. $$m=\frac{h}{e}⟹h=m\cdot e$$ (where $e=1.6\times 10^{-19}$ C is the charge of an electron).

• Intercept ($c$): The $y$-intercept ($-\frac{\varphi }{e}$) helps find the work function ($\varphi$) but is not directly needed for $h$.

Steps to find $h$:

1. Measure the slope ($m$) of the $V_o$ vs. $f$ graph.
2. Use $h=m\cdot e$ to calculate Planck’s constant.

Example Calculation (assuming slope $m=0.002$ V/Hz from the figure): $$h=0.002\times 1.6\times 10^{-19}=3.2\times 10^{-22}\text{ J⋅s}$$ (Note: The actual slope must be read from the graph in the exam paper.)

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iii) Effect of increasing frequency on stopping potential

As the frequency ($f$) of incident photons increases gradually:

• The kinetic energy of emitted photoelectrons ($K_{\text{max}}=hf-\varphi$) increases linearly.
• The stopping potential ($V_o$) must increase to halt the most energetic electrons.
• This is because $V_o$ is directly proportional to $K_{\text{max}}$ ($e{V}_o=K_{\text{max}}$).

Justification: From the equation $e{V}_o=hf-\varphi$, if $f$ increases, $V_o$ must rise to balance the equation. The graph shows a positive slope, confirming this relationship.

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b) Electron Beam in Electric and Magnetic Fields

i) Why the electron beam remains undeflected

An electron moving in crossed electric ($E$) and magnetic ($B$) fields experiences two forces:

1. Electric force ($F_E$): $F_E=eE$ (downward, if $E$ is upward).
2. Magnetic force ($F_B$): $F_B=evB$ (perpendicular to velocity and $B$).

For the beam to be undeflected, the two forces must cancel each other: $$F_E=F_B⟹eE=evB⟹E=vB$$ Here, the velocity ($v$) adjusts to balance the forces, making the net force zero.

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ii) Calculation of electron velocity

Given:

• Voltage ($V$) = 50 V (electric field $E=\frac{V}{d}$),
• Distance between plates ($d$) = 3 cm = 0.03 m,
• Magnetic field ($B$) = $10^{-4}$ T.

Step 1: Calculate electric field ($E$) $$E=\frac{V}{d}=\frac{50}{0.03}=1666.67\text{ V/m}$$

Step 2: Use the balance condition $E=vB$ $$v=\frac{E}{B}=\frac{1666.67}{10^{-4}}=1.6667\times 10^7\text{ m/s}$$

Final Answer: $$\boxed{v=1.67\times 10^7\text{ m/s}}$$

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OR Section (Radioactivity)

a) Definition of radioactivity

Radioactivity is the spontaneous emission of particles (α, β) or electromagnetic radiation (γ) from an unstable atomic nucleus to achieve stability. It is a random, exponential process governed by the decay constant ($\lambda$) and half-life ($t_{1/2}$).

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b) Comparison of decay rates after 18 hours

Given:

• Sample A: Half-life ($t_{1/2,A}$) = 3 hours,
• Sample B: Half-life ($t_{1/2,B}$) = 9 hours,
• Initial number of atoms: $N_0$ (equal for both).

Decay constant ($\lambda$) relation: $$\lambda =\frac{\ln 2}{t_{1/2}}$$

Decay rate ($R$) relation: $$R=\lambda N$$

After 18 hours:

1. Number of remaining atoms: $$N=N_0{e}^{-\lambda t}$$
2. Decay rates: $$R_A={\lambda }_A{N}_A,R_B={\lambda }_B{N}_B$$

Calculations:

• For Sample A: $${\lambda }_A=\frac{\ln 2}{3},N_A=N_0{e}^{-\frac{\ln 2}{3}\times 18}=N_0{e}^{-6\ln 2}=N_0{\left(\frac{1}{2}\right)}^6=\frac{N_0}{64}$$ $$R_A=\frac{\ln 2}{3}\times \frac{N_0}{64}=\frac{N_0\ln 2}{192}$$

• For Sample B: $${\lambda }_B=\frac{\ln 2}{9},N_B=N_0{e}^{-\frac{\ln 2}{9}\times 18}=N_0{e}^{-2\ln 2}=N_0{\left(\frac{1}{2}\right)}^2=\frac{N_0}{4}$$ $$R_B=\frac{\ln 2}{9}\times \frac{N_0}{4}=\frac{N_0\ln 2}{36}$$

Comparison: $$\frac{R_A}{R_B}=\frac{\frac{N_0\ln 2}{192}}{\frac{N_0\ln 2}{36}}=\frac{36}{192}=\frac{1}{5.33}$$

Conclusion: After 18 hours, Sample A decays faster than Sample B. Specifically: $$R_A:R_B=1:5.33$$

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c) Relation between half-life ($t_{1/2}$) and decay constant ($\lambda$)

The half-life is the time taken for half of the radioactive atoms to decay. Mathematically: $$N=\frac{N_0}{2}=N_0{e}^{-\lambda t_{1/2}}$$ Taking natural logs: $$\ln \left(\frac{1}{2}\right)=-\lambda t_{1/2}⟹\ln 2=\lambda t_{1/2}$$ Thus: $$\boxed{t_{1/2}=\frac{\ln 2}{\lambda }}$$

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d) Percentage of atoms decayed after five half-lives

After $n$ half-lives, the remaining fraction of atoms is: $$N=N_0{\left(\frac{1}{2}\right)}^n$$ For $n=5$: $$N=N_0{\left(\frac{1}{2}\right)}^5=\frac{N_0}{32}$$

Percentage decayed: $$\text{Decayed}=100\%-\left(\frac{1}{32}\times 100\%\right)=100\%-3.125\%=96.875\%$$

Final Answer: $$\boxed{96.875\%}$$