You place a chunk of naturally radioactive (it means not enriched for nuclear purposes) material on the not very exact scale and scale reads approximately 50 kg. You know that the half-life of this material is 20 days. After 40 days of observation the scale reading will be: Group of answer choices

approximately 50 kg

approximately 25 kg

approximately 12.5 kg

Between 12.5 kg and 25 kg, depending on activity

he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

Explanation:

A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression

              N = No

From this expression the quantity half life time () is defined with time so that half of the atoms have been transformed

            T_{1/2} = ln 2 /λ

in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei

first time of life

              N´ = ½ N

second time of life

              N´´ = ½ N´

              N´´ = ¼ N

consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is

             n_total = n₁ + n₂ + n₃

Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where

            m₂ < m₁

therefore mass of

             m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4

             m_total = m₂ ¾ N + m₁ ¼ N

             m_total = N (  ¾ m₂ + ¼ m₁)

Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y