Question

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

Answer 1

Answer:

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 $e^{- \alpha /t}$

 

From this expression the quantity half life time ($T_{1/2}$) 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.