The co-efficient of absorption for X-rays is related to the atomic number as

μ \propto \frac{1}{Z^{4}}

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μ \propto Z^{3}

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μ \propto \frac{1}{Z^{3}}

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μ \propto Z^{4}

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Solution

The correct option is C $μ \propto Z^{4}$
$μ = \frac{C N_{o} Z^{4} λ^{3}}{A}$

Hence, $μ \propto Z^{4}$

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Similar questions

Q. The intensity (1) of X-rays after traversing a distance x through a matter is related to the coefficient of absorption

(μ)

of the material as

If | $z_{1} | = | z_{2} | = | z_{3} | = | z_{4} | = 1$ and $z_{1} + z_{2} + z_{3} + z_{4} = 0$
then least value of the expression
$E = | z_{1} - z_{2} |^{2} + | z_{2} - z_{3} |^{2} + | z_{3} - z_{4} |^{2} + | z_{4} - z_{1} |^{2}$ is

Q. If

| z_{1} | = | z_{2} | = | z_{3} | = | z_{4} | = 1

and

z_{1} + z_{2} + z_{3} + z_{4} = 0

, then least value of the expression

E = | z_{1} - z_{2} |^{2} + | z_{2} - z_{3} |^{2} + | z_{3} - z_{4} |^{2} + | z_{4} - z_{1} |^{2}

Q. If

z_{1} + z_{2} + z_{3} + z_{4} = 0

, then the expression

| z_{1} - z_{2} |^{2} + | z_{2} - z_{3} |^{2} + | z_{3} - z_{4} |^{2} + | z_{4} - z_{1} |^{2} - 2 (| z_{1} |^{2} + | z_{2} |^{2} + | z_{3} |^{2} + | z_{4} |^{2})

is equal to

0

, is and only if,

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The co-efficient of absorption for X-rays is related to the atomic number as

The correct option is C μ∝Z4μ=CNoZ4λ3AHence, μ∝Z4

The correct option is C $μ \propto Z^{4}$
$μ = \frac{C N_{o} Z^{4} λ^{3}}{A}$

Hence, $μ \propto Z^{4}$