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Shielding with half-value layers

Used on the ASP, CSP and CHST. The worked example below is computed by code and checked against the source.

I = I0 × e^(−μx)

μ = 0.693 ÷ HVL

I0 = unshielded intensity; x = shield thickness; HVL = thickness that halves intensity (same units as x).

Worked example

80 mR/h, lead HVL 0.5 in, 1.5 in of lead.

80 × e^(−(0.693 ÷ 0.5) × 1.5) = 10.0 mR/h

1.5 in is three HVLs: 80 → 40 → 20 → 10.

Where it appears on the exam

Source

The formula comes from

OSHA, Ionizing Radiation: Introduction to Health Physics handout(opens osha.gov in a new tab)
“I = I o Be -µx = intensity on other side of shield ... Half-Value Layer (HVL): The shield thickness necessary to reduce intensity by half. HVL = 0.693 / µ.”

Questions

What is the formula for shielding with half-value layers?
I = I0 × e^(−μx); μ = 0.693 ÷ HVL. I0 = unshielded intensity; x = shield thickness; HVL = thickness that halves intensity (same units as x).
Will this formula be given on the BCSP exam?
BCSP says the formula is included in the question whenever one is needed, unless the question is testing whether you know the formula. You can't bring a formula sheet into the exam, so learn what each formula does and when to use it.

BCSP on formulas at the exam

BCSP, What to Expect When Sitting for a BCSP Exam (item 7)(opens bcsp.org in a new tab)
“For any question requiring a mathematical formula, the formula will be included within the exam question (unless the question is specifically testing your knowledge of a formula). You will not be provided or permitted to bring a formula sheet.”

Related

Checked 2026-10-16. Spot a problem? Email hello@passcited.com.

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