g Consider (12.5 A) micro-grams of a radioactive isotope with a mass number of (78 B) and a half-life of (32.6 C) million years. If energy released in each decay is 32.6 keV, determine the total energy released in joules (J) in 1 (one) year. Give your answer with three significant figures.

Energy released =  18.985 J

Explanation:

The  exponential decay of radioactive substance is given by -

N(t) = N₀

where

N₀ = initial quantity

k = decay constant

Half life,

⇒

Given,

N₀ = 12.5 + 3 = 15.5 × 10⁻⁶ gm

= 32.6 + 18 = 50.6 × 10⁶ years

So,

= 1.361 × 10⁻⁸ year⁻¹

Now,

N(1) = 15.5 × 10⁻⁶  

      = 15.49999978904

Now,

Substance decayed = N₀ - N(t)

                                 = 15.5 × 10⁻⁶ - 15.49999978904 × 10⁻⁶

                                 = 21.095 × 10⁻¹⁷ kg

⇒Δm = 21.095 × 10⁻¹⁷ kg

So,

Energy released = Δmc²

                           = 21.095 × 10⁻¹⁷ × 3 ×10⁸ × 3 × 10⁸

                          = 189.855 ×10⁻¹

                          = 18.985 J

⇒Energy released =  18.985 J

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

          Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is

Of F or Se, the larger atom is

Of Sn or I, the larger atom is

Of Cr or W, the larger atom is

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.