a) An oil drop of mass 3 10^ 11 g and of radius 2 10^ 4 cm carries 10 excess electrons. What is its terminal velocity when i) falling in a region in which there is no electric field? [1] ii) falling…

a) Terminal Velocity of Oil Drop

Given:

• Mass of oil drop, $m=3\times 10^{-11}\text{g}=3\times 10^{-14}\text{kg}$
• Radius, $r=2\times 10^{-4}\text{cm}=2\times 10^{-6}\text{m}$
• Excess electrons, $n=10$
• Viscosity of air, $\eta =1.8\times 10^{-5}{\text{N s m}}^{-2}$
• Electric field, $E=3\times 10^5{\text{V m}}^{-1}$ (downward)
• Charge of electron, $e=1.6\times 10^{-19}\text{C}$
• Acceleration due to gravity, $g=9.8{\text{m s}}^{-2}$

i) Terminal Velocity in Absence of Electric Field ($E=0$)

When the oil drop falls under gravity, it experiences:

• Gravitational force: $F_g=mg$
• Viscous drag (Stokes' law): $F_d=6\pi r\eta v_t$

At terminal velocity, $F_g=F_d$: $$mg=6\pi r\eta v_t$$ $$v_t=\frac{mg}{6\pi r\eta }$$

Calculations: $$m=3\times 10^{-14}\text{kg},r=2\times 10^{-6}\text{m}$$ $$v_t=\frac{(3\times 10^{-14})(9.8)}{6\pi (2\times 10^{-6})(1.8\times 10^{-5})}$$ $$v_t=\frac{2.94\times 10^{-13}}{6.75\times 10^{-10}\pi }$$ $$v_t=\frac{2.94\times 10^{-13}}{2.12\times 10^{-9}}\approx 1.39\times 10^{-4}{\text{m s}}^{-1}$$

Terminal velocity (no field): $v_t=1.39\times 10^{-4}{\text{m s}}^{-1}$ (downward)

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ii) Terminal Velocity in Electric Field ($E\ne 0$)

When an electric field is applied downward, the net force is: $$F_{\text{net}}=mg+neE-6\pi r\eta v_t$$ At terminal velocity, $F_{\text{net}}=0$: $$mg+neE=6\pi r\eta v_t$$ $$v_t=\frac{mg+neE}{6\pi r\eta }$$

Calculations: $$ne=10\times 1.6\times 10^{-19}=1.6\times 10^{-18}\text{C}$$ $$neE=(1.6\times 10^{-18})(3\times 10^5)=4.8\times 10^{-13}\text{N}$$ $$mg=2.94\times 10^{-13}\text{N}$$ $$v_t=\frac{(2.94+4.8)\times 10^{-13}}{2.12\times 10^{-9}}=\frac{7.74\times 10^{-13}}{2.12\times 10^{-9}}\approx 3.65\times 10^{-4}{\text{m s}}^{-1}$$

Terminal velocity (with field): $v_t=3.65\times 10^{-4}{\text{m s}}^{-1}$ (downward)

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b) Thomson’s Method: Perpendicular $E$ and $B$ Fields

In Thomson’s experiment to determine $\frac{e}{m}$ of an electron, the electric ($E$) and magnetic ($B$) fields are kept perpendicular for the following reasons:

Mathematical Explanation: For equilibrium (no deflection), $$eE=evB⟹E=vB$$ This condition is only possible if $E\perp B$, ensuring the electron moves in a straight line (no net force) or a parabola (when fields are adjusted for deflection).

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c) Use of X-rays in Millikan’s Oil Drop Experiment

X-rays are used in Millikan’s experiment to ionize air molecules, producing free electrons that attach to oil droplets. This allows:

1. Controlled Charging: Oil drops acquire a known number of electrons ($n$), enabling precise measurement of charge.
2. Repeatability: Ensures droplets have discrete charges (multiples of $e$), confirming quantization of charge.
3. Stability: Ionized air helps maintain a consistent electric field for accurate terminal velocity calculations.

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OR

a) Decay Constant ($\lambda$) and Half-Life ($t_{1/2}$)

• Decay Constant ($\lambda$): The probability per unit time that a radioactive nucleus will decay. It is related to the disintegration rate and has units of s⁻¹. $$\lambda =\frac{\text{Number of decays per second}}{\text{Total number of nuclei}}$$

• Half-Life ($t_{1/2}$): The time required for half of the radioactive nuclei in a sample to decay. It is inversely proportional to $\lambda$: $$t_{1/2}=\frac{\ln 2}{\lambda }$$

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b) Derivation of $N=N_0{e}^{-\lambda t}$

Assumptions:

• A sample has $N_0$ radioactive nuclei at $t=0$.
• The decay rate is proportional to the number of undecayed nuclei.

Step 1: Define Decay Rate The rate of decay ($-\frac{dN}{dt}$) is proportional to $N$: $$-\frac{dN}{dt}=\lambda N$$

Step 2: Solve the Differential Equation Separate variables and integrate: $$\frac{dN}{N}=-\lambda dt$$ $${\int }_{N_0}^N\frac{dN}{N}=-\lambda {\int }_0^{t}dt$$ $$\ln N-\ln N_0=-\lambda t$$ $$\ln \left(\frac{N}{N_0}\right)=-\lambda t$$

Step 3: Exponentiate to Solve for $N$ $$\frac{N}{N_0}=e^{-\lambda t}$$ $$N=N_0{e}^{-\lambda t}$$

Interpretation:

• $N$ decreases exponentially with time.
• $\lambda$ determines the decay rate.
• At $t=t_{1/2}$, $N=\frac{N_0}{2}$, confirming the half-life relation.