a A radioactive sample has a half life of 8.3x10^4 years. Calculate its disintegration constant and time taken for 25% of its activity to disappear.
Answers 1
Answer:
It will take 38298.5 years for 25% of its activity to disappear.
Explanation:
Given the half life = 8.3x10⁴ years = [tex]t_{\frac{1}{2} }[/tex]
We know that λ (disintegration constant) = log 2 / [tex]t_{\frac{1}{2} }[/tex]
=> λ = log 2 / 8.3x10⁴
=> λ = 3.62 × 10⁻⁶
We know that λ = [tex]\frac{2.303}{t}[/tex] log [tex]\frac{N_{0} }{N}[/tex]
where N₀ = initial number of atoms
and N = final number of atoms
Here, N₀ = N₀ and N = 25%*N₀ = N₀/4
Let the time required for 25% of its activity to disappear be t
=> 3.62 × 10⁻⁶ = [tex]\frac{2.303}{t}[/tex] * log [tex]\frac{N_{0} }{\frac{N_{0}}{4} }[/tex]
=> 3.62 × 10⁻⁶ = [tex]\frac{2.303}{t}[/tex] * log [tex]\frac{4* N_{0} }{N_{0}}}[/tex]
=> 3.62 × 10⁻⁶ = [tex]\frac{2.303}{t}[/tex] * log 4
=> 3.62 × 10⁻⁶ = [tex]\frac{2.303}{t}[/tex] * 0.602
=> 6.01 × 10⁻⁵ = [tex]\frac{2.303}{t}[/tex]
=> t = 38298.5 years.
Therefore, it will take 38298.5 years for 25% of its activity to disappear.
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mitchell170
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Given data : A man wants to reach point B on the opposite bank of a river flowing at a speed 4 m/s.
Solution : Firstly we have to find out, distance between A to B.
Now, according to figure,
➜ Distance AC = 40 m
➜ Distance BC = 30 m
➜ ∠ACB = 90⁰
Now, by Pythagoras theorem,
➜ {AB}² = {AC}² + {BC}²
➜ {AB}² = {40}² + {30}²
➜ {AB}² = 1600 + 900
➜ {AB}² = 2500
➜ AB = √2500
➜ AB = 50 m
Case 1 : If man move A directly to B, it is not possible because of the speed of the river. When a man matches his speed with the speed of the river, he will reach point B.
Let, speed of man is equal to speed of river.
Hence,
➜ Speed of man = Distance/Time
➜ 4 = AB/Time
➜ 4 = 50/Time
➜ Time = 50/4 ----{1}
Case 2 : If man move A directly to B, it is not possible because of the speed of the river, So man needs to walk in vertical direction to A (which is A to C), due to presence of river motion he will reach to point B.
Let, minimum speed of man be x.
Hence,
➜ Speed of man = Distance/Time
➜ x = AC/Time
➜ x = 40/Time
➜ Time = 40/x ----{2}
According to case 1 and case 2 the man reaches at point B. So,
Assume that man take same time in both cases to reach at point B.
Hence, from eq. {1} and eq. {2}
➜ 50/4 = 40/x
➜ x = {40 * 4}/50
➜ x = 160/50
➜ x = 16/5 m/s
Answer : Hence, the minimum speed of man is 16/5 m/s.
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