A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:
- 3.75 hours
- 0.56 hours
- 0.26 hours
- 1.80 hours
The Correct Option is D
Solution and Explanation
The correct answer is (D): 1.80 hours
Explanation:
\(\frac{dA }{dt }\)- (-λ1A) + ( -λ2A)
⇒ \(\frac{dA }{dt }\) = - ( λ1 + λ2 )A
⇒ λeff = λ1 + λ2
⇒ \(\frac{ln2}{(t_{1/2})eff }\) = \(\frac{ln2}{(t_{1/2})_1}\) + \(\frac{ln2}{(t_{1/2})_2}\)
⇒ \(( t_{1/2} )\)eff =\(\frac{ 4.5 × 3 }{7.5}\) hours
= 1.8 hours
Top Questions on Nuclei
- Derive an expression for law of radioactive decay. Define one becquerel (Bq).
- Derive an expression for law of radioactive decay. Define one becquerel (Bq).
- What is the meaning of nucleon? What is the mass number of nucleus? Write down the relation between mass number and radius of nucleus and show that the density of nucleus does not depend on mass number.
- What is meant by mass-defect? In a nuclear fission reaction, find out the value of the fission energy \( Q \).
- Explain Nuclear fusion.
Questions Asked in JEE Main exam
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
- JEE Main - 2025
- Differential Calculus
- A 400 g solid cube having an edge of length \(10\) cm floats in water. How much volume of the cube is outside the water? (Given: density of water = \(1000 { kg/m}^3\))
- An infinite wire has a circular bend of radius \( a \), and carrying a current \( I \) as shown in the figure. The magnitude of the magnetic field at the origin \( O \) of the arc is given by:
- JEE Main - 2025
- Magnetic Field
- Let \( A = [a_{ij}] \) be a matrix of order 3 \(\times\) 3, with \(a_{ij} = (\sqrt{2})^{i+j}\). If the sum of all the elements in the third row of \( A^2 \) is \( \alpha + \beta\sqrt{2} \), where \(\alpha, \beta \in \mathbb{Z}\), then \(\alpha + \beta\) is equal to:
- JEE Main - 2025
- Matrices and Determinants
- Total number of non-bonded electrons present in \( \text{NO}_2 \); ion based on Lewis theory is:
- JEE Main - 2025
- Chemical bonding and molecular structure
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