In spectrophotometry, which expression correctly relates absorbance to transmittance when %T is used?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards
From $24.99Unlock all

Enhance your knowledge with the Ciulla Clinical Chemistry Test. Study with interactive quizzes, featuring flashcards and multiple-choice questions. Each question offers hints and explanations for better understanding. Prepare effectively and excel in your exam!

Multiple Choice

In spectrophotometry, which expression correctly relates absorbance to transmittance when %T is used?

Explanation:
Absorbance grows as transmittance drops, and its relation to light transmission is A = -log10(T), where T is the fraction of light that passes through (0 < T ≤ 1). If you’re using percent transmittance, %T, then T = %T/100. Substituting gives A = -log10(%T/100) = -log10(%T) + log10(100) = 2 − log10(%T). So the correct expression is A = 2 − log10(%T). For example, if %T = 50, then A ≈ 2 − log10(50) ≈ 0.301, which matches A = -log10(0.5). The other forms fail because they omit the negative sign, or they don’t account for dividing by 100 inside the logarithm.

Absorbance grows as transmittance drops, and its relation to light transmission is A = -log10(T), where T is the fraction of light that passes through (0 < T ≤ 1). If you’re using percent transmittance, %T, then T = %T/100. Substituting gives A = -log10(%T/100) = -log10(%T) + log10(100) = 2 − log10(%T). So the correct expression is A = 2 − log10(%T).

For example, if %T = 50, then A ≈ 2 − log10(50) ≈ 0.301, which matches A = -log10(0.5). The other forms fail because they omit the negative sign, or they don’t account for dividing by 100 inside the logarithm.

More Multiple Choice Questions
Subscribe

Get the latest from Examzify

You can unsubscribe at any time. Read our privacy policy