The half-life of cesium-137 is 30 years. Suppose that we start with 70 grams of cesium-137 in a storage pool. How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool?

Explanation:

Let n be number of half lives taken .

10 = 70 x ( 1/2 )ⁿ

1/7 = ( 1/2 )ⁿ

2ⁿ = 7

n ln2 = ln 7

n = ln7 / ln2

= 1.9459 / .693

= 2.8

So 2.8 half lives will be required .

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.