During a negative beta decay
- an atomic electron is ejected
- an electron which is already present within the nucleus is ejected
- a neutron in the nucleus decays emitting an electron
- a part of the binding energy of the nucleus is converted into an electron
The Correct Option is C
Solution and Explanation
$_0n^1\rightarrow_1H^1+_{-1}e^0+v$
v is antineutrino.
Top Questions on deccay rate
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Questions Asked in JEE Advanced exam
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
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The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
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Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
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Concepts Used:
Decay Rate
The disintegration of unstable heavy atomic nuclei into lighter, more stable, atomic nuclei, accompanied in the process by the emission of ionizing radiation (alpha particles, beta particles or gamma rays). This is a random process at the atomic level but, given a large number of similar atoms, the decay rate on average is predictable, and is usually measured by the half-life of the substance.
The equation for finding out the decay rate is given below:
