Question

Calculate the mass pb-214 which has an activity of 2 curie, if it's half life is 26.8minutes

Answer 1

The mass of N atoms is m =  3.064 × 10^−8 g

Explanation:

Here T=26.8 minutes  = 26.8 × 60 sec

Decay constant

λ = 0.693 T = 0.69326.8 × 60 = 4.32 × 10^4 sec−1

Now 1 curie is equal to 3.71 × 10^10 disintegrations per second

λ = 3.71×10^10

If N be the number of atoms in one curie, then

−dN / dt = λN

3.71×1010=431×10−4  

N = 3.71 × 10^10 / 4.31 × 10^−4

N =8.607 × 10^13

Further, atomic weight of RaB = 214 and Avogadro's number = 6.025 × 10^23

Mass of one atom = 214 / 6.025 × 10^23

Mass of N atoms  = (2146.025 × 10^23) × (8.607 × 10^13)

Mass of N atoms  = 3.064 × 10^−8 g

Thus the mass of N atoms is m =  3.064 × 10^−8 g

Also learn more

The mass of an atom of an element x is 2 × 10^-23 grams calculate its atomic mass ?

https://brainly.in/question/6906227

Answer 2

Answer:

The mass of Pb-214 is $m = 6.10 \times 10^{-8} gram$

Explanation:

As we know that the activity of the radioactive substance is given as

$A = N\lambda$

given that

$A = 2 Ci$

$A = 2\times 3.7 \times 10^{10}$

$A = 7.4 \times 10^{10} Bq$

$7.4 \times 10^{10} = N(\frac{ln2}{26.8 \times 60})$

$N = 1.72 \times 10^{14}$

now number of moles of Pb is given as

$n = \frac{N}{N_a}$

$n = \frac{1.72\times 10^{14}}{6.02\times 10^23}$

$n = 2.85 \times 10^{-10}$

now the mass of the Pb

$m = 214\times 2.85 \times 10^{-10}$

$m = 6.10 \times 10^{-8} gram$

#Learn

Topic : Radioactivity

https://brainly.in/question/8246714