The correct answer is (C) : 7.5 years
\(∵A=\frac{A0}{\frac{t}{2^{T_{1/2}}}}\)
\(⇒2^{\frac{t}{T_{1/2}}}\)
\(=\frac{A_0}{A}=16\)
\(⇒\frac{t}{T_{1/2}}=4\)
\(⇒\frac{30}{T_{1/2}}=4\)
\(⇒T_{1/2}=\frac{30}{4}\)
=7.5 years
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Radioactivity is a phenomenon observed in certain elements where unstable atomic nuclei spontaneously emit energy and subatomic particles. This process is driven by the desire of the nucleus to achieve a more stable state. It's crucial to understand the three main types of radioactive decay:
Alpha Decay: In alpha decay, a nucleus emits an alpha particle, consisting of two protons and two neutrons.
Beta Decay: Beta decay involves the emission of a beta particle, which can be a positron or an electron, from an unstable nucleus.
Gamma Decay: Gamma decay releases gamma rays, electromagnetic radiation, to achieve a more stable nuclear state.
The emission of these particles and energy is a result of nuclear instability. The rate of decay is characterized by the half-life, the time taken for half of the radioactive material to undergo decay. Radioactivity has diverse applications, from medical treatments and industrial processes to power generation in nuclear reactors.