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State two practical uses of photo cells.

Physics

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(b) (i) State two practical uses of photo cells.

  1. Automatic street lighting: Photo cells detect ambient light levels and switch streetlights on at dusk and off at dawn.
  2. Automatic door openers: They can detect the presence of a person by sensing an interruption in a light beam, triggering the door to open. Other uses include burglar alarms, light meters in cameras, and solar panels for electricity generation.

(b) (iii) A scientist uses a radioactive source with a half-life of 4 hrs4 \, \text{hrs} as a clock. If the source initially emits 1220 particles per minute, what time interval has elapsed when he records 160 counts per minute?

  1. Identify the given values.

    Initial activity (N0N_0) = 1220 counts/min
    Final activity (NN) = 160 counts/min
    Half-life (T1/2T_{1/2}) = 4 hrs4 \, \text{hrs}

  2. Use the radioactive decay formula to find the number of half-lives (nn).

    The formula for radioactive decay is:

    N=N0(12)nN = N_0 \left(\frac{1}{2}\right)^n

    Substitute the given values:

    160=1220(12)n160 = 1220 \left(\frac{1}{2}\right)^n

    Divide both sides by 1220:

    1601220=(12)n\frac{160}{1220} = \left(\frac{1}{2}\right)^n 0.1311475≈(12)n0.1311475 \approx \left(\frac{1}{2}\right)^n

    Take the natural logarithm of both sides:

    ln⁡(0.1311475)=nln⁡(12)\ln(0.1311475) = n \ln\left(\frac{1}{2}\right) −2.0316≈n(−0.6931)-2.0316 \approx n (-0.6931)

    Solve for nn:

    n≈−2.0316−0.6931n \approx \frac{-2.0316}{-0.6931} n≈2.931n \approx 2.931
  3. Calculate the total time elapsed (tt).

    The total time elapsed is the number of half-lives multiplied by the half-life period:

    t=n×T1/2t = n \times T_{1/2} t≈2.931×4 hrst \approx 2.931 \times 4 \, \text{hrs} t≈11.724 hrst \approx 11.724 \, \text{hrs}

    The time interval elapsed is 11.72 hrs\boxed{\text{}11.72 \, \text{hrs}}.

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