A radioactive nucleus is being produced at a constant rate $\alpha$ per second. Its decay constant is $\lambda$. If $N_0$ are the number of nuclei at time t = 0, then the maximum number of nuclei possible are
- $N_{0}+\frac{\alpha}{\lambda}$
- $N_0$
- $\frac{\lambda}{\alpha}+N_{0}$
- $\frac{\alpha }{\lambda }$
The Correct Option is D
Approach Solution - 1
Approach Solution -2
Here, dn/dt is quantified in units called Becquerels and the source of the term is the number of disintegrations per second.
dn/dt = α, according to the given problem.
dn/dt is also equal to Nλ.
Therefore, α = Nλ
When the rate of decay equals the rate of generation of new nuclei, or at steady state, the maximum number of nuclei will be present. Therefore, the greatest number of nuclei will be present in this scenario.
So, N = α/λ
The number of nuclei undergoing decay and the rate of decay per unit of time is directly proportional to the total number of nuclei present in the given sample of the radioactive material, according to the law of radioactive decay, whenever the material experiences a decay (alpha, beta, or gamma decay).
Mathematical representation of the Law of radioactive decay:
Where,
ΔN = number of nuclei in the sample that undergoes radioactive decay.
N = the total number of nuclei in the sample
Δt =unit time
Simplifying the equation: ΔN = λN. Δt
Where λ is the radioactive decay constant.
Top Questions on Nuclei
- The disintegration energy \( Q \) for the nuclear fission of \( ^{235}\text{U} \to ^{140}\text{Ce} + ^{94}\text{Zr} + n \) is _____ MeV.
Given atomic masses:
\[^{235}\text{U} = 235.0439 \, u, \quad ^{140}\text{Ce} = 139.9054 \, u, \quad ^{94}\text{Zr} = 93.9063 \, u, \quad n = 1.0086 \, u\]Value of \( c^2 = 931 \, \text{MeV/u} \). - The mass defect in a particular reaction is \( 0.4 \, \text{g} \). The amount of energy liberated is \( n \times 10^7 \, \text{kWh} \), where \( n = \) _____. (speed of light \( = 3 \times 10^8 \, \text{m/s} \))
- A nucleus has mass number \(A_1\) and volume \(V_1\). Another nucleus has mass number \(A_2\) and volume \(V_2\). If the relation between mass numbers is \(A_2 = 4A_1\), then \(\frac{V_2}{V_1} =\) _______.
- The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:
- A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2:1. After disintegration they will move
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- Transcription
Concepts Used:
Nuclei
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
Read More: Nuclei
Following are the terms related to nucleus:
- Atomic Number
- Mass Number
- Nuclear Size
- Nuclear Density
- Atomic Mass Unit