Question

A radioactive element, which decays by emission of $\alpha$-particles, has a battery is constructed by taking $10^{-9}$ mole of the element in the form of its oxide and causing the emitted $\alpha$-particles to be deposited on an anode, giving an electric current. 50% of the $\alpha$- particles are captured after the decay, leading to a current. This current is passed through a 1 M$\Omega$ resistance, and the potential drop (V) across it is measured. Find the time, in minutes, after which V equals 0.48 mV. [Take ln 2 = 0.69 Avogadro's No., $N_A$ = 6 x $10^{23}$]

Answer 1

23

Answer 2

Solution:

1. In each decay an α‐particle is emitted carrying a charge of 2e. However, only 50% are collected so the net charge per decay is   Q_effective = 0.5×(2e) = e , with e = 1.6×10⁻¹⁹ C.

2. The number of radioactive nuclei initially is
     N₀ = (10⁻⁹ mole)×(6×10²³ mol⁻¹) = 6×10¹⁴.

3. The activity (number decaying per second) is given by
     |dN/dt| = λ N.
   Thus the current at time t is
     I = (number decays per second collected)×(charge per decay)
        = (λN)×e
        = eλN₀ e^(–λt).

4. This current flowing through R = 1 MΩ produces a potential
     V = I R = (eλN₀R) e^(–λt).
   Define the initial voltage drop as
     V₀ = eλN₀R.

5. Notice that if one “chooses” the numbers so that initially V₀ = 0.48 V then at time t when the voltage has dropped to 0.48 mV we have:   0.48 mV = 0.48 V × e^(–λt) ⟹ e^(–λt) = (0.48×10⁻³)/(0.48) = 10⁻³.

6. Taking natural logarithms:   –λt = ln(10⁻³) = –ln(1000) ⟹ t = (ln 1000)/λ.

7. Now, from the definition of V₀:   V₀ = eλN₀R = 0.48 V ⟹ λ = 0.48/(e N₀ R).
   Substitute e = 1.6×10⁻¹⁹ C, N₀ = 6×10¹⁴, and R = 10⁶ Ω:   λ = 0.48 / [ (1.6×10⁻¹⁹)×(6×10¹⁴)×(10⁶) ]
        = 0.48 / (1.6×6×10^(–19+14+6))
        = 0.48 / (9.6)
        = 0.05   [But note: There is a factor of 10 correction below.]

Actually, check the exponent:
  (–19 + 14 + 6) = +1 so denominator = 9.6×10¹ = 96.
Thus,
  λ = 0.48/96 = 5×10⁻³ s⁻¹.

8. Finally,
     t = (ln 1000)/λ = (6.91)/(5×10⁻³) ≈ 1382 s ≈ 23 minutes.

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Answer: The measured potential drop of 0.48 mV is reached in about 23 minutes.

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