How many electrons, protons and neutrons are there in 12 gm of C 12 6 ? Avogadro's number = 6 . 022 × 10 23 at 1 gm at wt. How many electrons, protons and . neutrons are there in 12 gm of .C'? ?How many electrons, protons and . neutrons are there in 12 gm of .C'? ?How many electrons, protons and . neutrons are there in 12 gm of .C'? ?

In C 12 6 , number of protons are = 6 , number of neutrons are = 6 , number of electrons are = 6 . 12 gm of C 12 6 = 1 mole So the number of protons in 12 gm of C 12 6 is = 6 × 6 × 10 23 = 36 × 10 23 . So the number of neutrons in 12 gm of C 12 6 is = 6 × 6 × 10 23 = 36 × 10 23 . So the number of electrons in 12 gm of C 12 6 is = 6 × 6 × 10 23 = 36 × 10 23 .

0 . 1 gm atom of radio-active isotope X A Z (half-life 5 days) is taken. How many number of atoms will decay during eleventh day.

As per question The amount left after 10 days = amount going to disintegrate on 11 th day Here, initial amount of radio-active isotope, N ∘ = 0 . 1 gm half life period, t 1 2 = 5 days decay time, t = 10 days Integrated rate law expression for 1 st order decay process : 0 . 693 t 1 2 = 2 . 303 t log N ∘ N ⇒ 0 . 693 5 = 2 . 303 10 log 0 . 1 N 10 , here N 10 is the amount left after 10 days ⇒ 0 . 6018 = log 0 . 1 N 10 ⇒ N 10 = 0 . 0250 gm atom Similarly, for the decay period, t = 11 days 0 . 693 5 = 2 . 303 11 log 0 . 1 N 11 , here N 11 is the amount left after 11 days ⇒ 0 . 662 = log 0 . 1 N 11 ⇒ N 11 = 0 . 0218 gm atom So, the amount decayed in 11 th day : N 10 - N 11 = 0 . 250 - 0 . 0218 = 3 . 2 × 10 - 3 gm atom . N 10 - N 11 = 3 . 2 × 10 - 3 gm atom = 3 . 2 × 10 - 3 × 6 . 023 × 10 23 atoms = 1 . 93 × 10 21 atoms . Hence, 1 . 93 × 10 21 number of atoms will decay during eleventh day.

The half-life of radon is 3 . 8 days . After how many days will 1 20 th of radon sample be left over?

Given, half-life of radon : t 1 2 = 3 . 8 days . Using the relation between half life time ( t 1 2 ) and decay constant ( λ ) : λ = 0 . 693 t 1 2 = 0 . 693 3 . 8 = 0 . 182 day - 1 Let, initial amount of radon be N ∘ and after t days the amount is N . Given, N = N ∘ 20 Then using equation : t = 2 . 303 λ log 10 N ∘ N ⇒ t = 2 . 303 0 . 182 log 10 N ∘ N ∘ 2 ⇒ t = 2 . 303 0 . 182 log 10 20 = 16 . 54 days Hence, after 16 . 54 days 1 20 th of radon sample be left over.

What is the activity of one gram of Ra 86 236 whose half-life is 1622 years?

The activity of a sample of radioactive matter is defined by the number of disintegrations taking place at its core at any given moment. And one curie is equal to that quantity of radioactive material in which the number of atoms decaying per second is equal to 37 billion ( 3 . 7 × 10 10 ) ). It was based on the rate of decay of atoms within one gram of radium in one half life. Hence, required activity of Ra 86 236 is 1 Curie i.e. 3 . 7 × 10 10 disintegrations per second.

If 3 × 10 - 9 Kg of Au 79 200 has an activity of 58 . 9 ci , what is its half-life?

N = 3 × 10 - 9 × 6 . 023 × 10 23 200 = 9 × 10 15 atoms Activity R = 58 . 9 Curie = 58 . 9 × 3 . 7 × 10 10 dps = 21 . 8 × 10 12 dps R = λ N , where λ is decay constant λ = 0 . 693 t 1 2 ⇒ R = 0 . 693 t 1 2 × N ⇒ t 1 2 = 0 . 693 N R ⇒ t 1 2 = 0 . 693 × 9 × 10 15 21 . 8 × 10 12 = 2 . 88 × 10 3 s Hence, half life is 2 . 88 × 10 3 seconds.