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Th6 intensity of $γ$ radiation from a given source is I. On passing through 36 mm of lead, it is reduced to I / 8. The thickness of lead, which will reduced the intensity to I/ 2will be

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Correct option is C

we know that I′=Ie−μx or −μX=log⁡I′/I −μ(36)=log⁡(I/8I) −μx′=log⁡(I/2I) Dividingeq.(1)byeq.(2)363lo8(t 36x′=3log⁡(1/2)log⁡(1/2)=3 x′=363=12mm

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Th6 intensity of γ radiation from a given source is I. On passing through 36 mm of lead, it is reduced to I / 8. The thickness of lead, which will reduced the intensity to I/ 2will be

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Th6 intensity of $γ$ radiation from a given source is I. On passing through 36 mm of lead, it is reduced to I / 8. The thickness of lead, which will reduced the intensity to I/ 2will be