Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively . If they have the same activity at certain instant t, what is the ratio of the number of atoms of S1 to S2 at instant t?
Answers
Let the Sample S₁ contains (before Decay) : N₁ atoms
Let the Sample S₂ contains (before Decay) : N₂ atoms
We know that Decay Rate(R) of Sample is given by :
R = λN { where λ Radioactive Decay constant = 0.693/T(Half) }
Given that at a Certain instant 't' the Sample's S₁ and S₂ have same activity (i.e. Decay rate)
⇒ At that instant 't' : Rate of Decay of Sample - 1 = Rate of Decay of Sample - 2
Rate of Decay of Sample - 1 (R₁) = λN = (0.693/T(Half of S₁) × N₁ )
⇒ R₁ = {0.693/(3 × 60 × 60)} × N₁
Rate of Decay of Sample - 2 (R₂) = λN = (0.693/T(Half of S₂) × N₂ )
⇒ R₂ = {0.693/(7 × 60 × 60)} × N₂
At Time 't' : R₁ = R₂
⇒ {0.693/(3 × 60 × 60)} × N₁ = {0.693/(7 × 60 × 60)} × N₂
⇒ N₁/3 = N₂/7
⇒ N₁/N₂ = 3/7
⇒ The Ratio of the Number of atoms of S₁ and S₂ at instant 't' is 3 : 7
The answer is 3:7
GIVEN
Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively.
TO FIND
the ratio of the number of atoms of S1 to S2 at instant t?
SOLUTION
We can simply solve the above problem as follows;
Let the number of atoms in S1 before decay = N₁
The Number of atoms in S2 before decay = N₂
Decay rate (R) of a sample is given by the formula;
R = λN
Where,
λ = Radioactive Decay constant = 0.693/T(Half)
Given, At an instant 't' the samples have same activity.
Therefore,
At that instant 't':
Rate of Decay of Sample - 1 = Rate of Decay of Sample - 2
Rate of Decay of Sample - 1 (R₁) = AN = (0.693/T(Half of S₁) × N₁)
R₁ = {0.693 /(3×60×60)} × N₁
R₂ = {0.693/(7 × 60 × 60)} × N₂
GIVEN,
At 't',
R₁ = R₂
= {0.693 /(3×60×60)} × N₁ = {0.693/(7 × 60 × 60)} × N₂
N₁/3 = N₂/7
N₁/N₂ = 3/7 = 3:7
Hence, The answer is 3:7
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