Nuclei of a radioactive element are being produced at a constant rate - The element has a decay constant - At time t -0- there are N 0 nuclei of the element- If -2 N 0- the number N of nuclei as t - isA- N 0-2B- N 0-4C- 2 N 0D- 4 N 0

The correct option is C 2N_0 If α = 2 N_0 λ, then we have n=1/λ[2N_0 λ (2N_0 λ λ N_0)e^ λ t] =N_0 (2 e^ λ t) When t →∞ N=N_0 (2 e^ )= N_0 (2 0)=2N_0 ...

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Byju's Answer

Standard XII

Physics

Beta Decay

Nuclei of a r...

Question

Nuclei of a radioactive element are being produced at a constant rate $α$ . The element has a decay constant $λ$ . At time t = 0, there are $N_{0}$ nuclei of the element.

If $α = 2 N_{0} λ$ , the number N of nuclei as t $\to \infty$ is

$\frac{N_{0}}{4}$

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$\frac{N_{0}}{2}$

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$2 N_{0}$

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$4 N_{0}$

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Solution

The correct option is C
$2 N_{0}$

If $α = 2 N_{0} λ$ , then we have

$n = \frac{1}{λ} [2 N_{0} λ - (2 N_{0} λ - λ N_{0}) e^{- λ t}]$

$= N_{0} (2 - e^{- λ t})$

When t $\to \infty$

$N = N_{0} (2 - e^{-}) = N_{0} (2 - 0) = 2 N_{0}$

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Nuclei of a radioactive element are being produced at a constant rate α. The element has a decay constant λ. At time t = 0, there are N0 nuclei of the element. If α=2N0λ, the number N of nuclei as t →∞ is

The correct option is C 2N0 If α=2N0λ, then we have n=1λ[2N0λ−(2N0λ−λN0)e−λt] =N0(2−e−λt) When t →∞ N=N0(2−e−)=N0(2−0)=2N0

Nuclei of a radioactive element are being produced at a constant rate $α$ . The element has a decay constant $λ$ . At time t = 0, there are $N_{0}$ nuclei of the element.

If $α = 2 N_{0} λ$ , the number N of nuclei as t $\to \infty$ is

The correct option is C
$2 N_{0}$

If $α = 2 N_{0} λ$ , then we have

$n = \frac{1}{λ} [2 N_{0} λ - (2 N_{0} λ - λ N_{0}) e^{- λ t}]$

$= N_{0} (2 - e^{- λ t})$

When t $\to \infty$

$N = N_{0} (2 - e^{-}) = N_{0} (2 - 0) = 2 N_{0}$