A radioactive material, with half-life of six months, has 100 thousand unstable nuclei. (a) Find the number of unstable nuclei present after three months.
(b) What will be its activity at this time (after three months)?

See Explanation

Explanation:

Given that;

N/No = (1/2)^t/t1/2

Where;

No = amount of radioactive isotope originally present

N = A mount of radioactive isotope present at time t

t = time taken

t1/2 = half life

N/1000=(1/2)^3/6

N/1000=(1/2)^0.5

N = (1/2)^0.5 * 1000

N= 707 unstable nuclei

Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;

Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1

Hence;

A=Aoe^-kt

Where;

A = Activity after a time t

Ao = initial activity

k = decay constant

t = time taken

A = Aoe^-3 *0.1155

A=Aoe^-0.3465

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

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