Final answer:
To find the minority carrier lifetime in a Haynes-Shockley experiment where the amplitudes differ by a factor of 10 at time intervals 25μs and 100μs, one can use the formula τ = (t2 - t1) / ln(10), which yields a lifetime of approximately 32.5 microseconds.
Step-by-step explanation:
In the Haynes-Shockley experiment, which measures the properties of minority carriers in semiconductors, the minority carrier lifetime (τ) can be estimated from the decay of the amplitude of the carriers over time. According to the experiment parameters provided, the maximum amplitudes at times t1 (25μs) and t2 (100μs) differ by a factor of 10. This relates to the exponential decay of the carrier density over time which can be modeled by N(t) = N0e-t/τ, where N(t) is the number of carriers at time t, and N0 is the initial number of carriers.
To find the minority carrier lifetime, we can set up a ratio of the amplitudes at t1 and t2 times and solve for τ:
τ = (t2 - t1) / ln(10)
Plugging in the given times:
τ = (100μs - 25μs) / ln(10)
τ = 75μs / ln(10)
τ ≈ 32.5μs
Therefore, the minority carrier lifetime is approximately 32.5 microseconds.