Question:
n/p ratio during positron decay:
n/p ratio during positron decay:
Show Hint
\(\beta^+\) decay: \(p \rightarrow n\) \(\Rightarrow\) \(n\) increases, \(p\) decreases \(\Rightarrow\) \(n/p\) ratio increases.
Updated On: Apr 16, 2026
- increases
- decreases
- remains constant
- All of these
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Positron emission (\(\beta^+\) decay) is a type of radioactive decay in which a proton inside the nucleus is converted into a neutron, a positron (\(e^+\)), and a neutrino (\(\nu\)):
\[
p \rightarrow n + e^+ + \nu
\]
Step 1: Effect on proton and neutron numbers.
• Number of protons (\(Z\)) decreases by 1
• Number of neutrons (\(N\)) increases by 1
• Mass number (\(A = N + Z\)) remains constant
Step 2: Effect on \(n/p\) ratio. Initial \(n/p\) ratio = \(\frac{N}{Z}\). After decay: \[ n' = N + 1,\quad p' = Z - 1 \] \[ \frac{n'}{p'} = \frac{N+1}{Z-1}>\frac{N}{Z} \] The \(n/p\) ratio increases because the number of neutrons increases while the number of protons decreases.
Step 3: Conclusion. During positron decay, the \(n/p\) ratio increases.
Step 1: Effect on proton and neutron numbers.
• Number of protons (\(Z\)) decreases by 1
• Number of neutrons (\(N\)) increases by 1
• Mass number (\(A = N + Z\)) remains constant
Step 2: Effect on \(n/p\) ratio. Initial \(n/p\) ratio = \(\frac{N}{Z}\). After decay: \[ n' = N + 1,\quad p' = Z - 1 \] \[ \frac{n'}{p'} = \frac{N+1}{Z-1}>\frac{N}{Z} \] The \(n/p\) ratio increases because the number of neutrons increases while the number of protons decreases.
Step 3: Conclusion. During positron decay, the \(n/p\) ratio increases.
Was this answer helpful?
0
0
Top MET Chemistry Questions
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
- Ethyl chloride is converted into diethyl ether by:
- Correct order of reactivity:
- A group of 13 element if added in small amounts to Ge, then the type of semiconductor formed is
- Consider the following statements I. Repulsive forces are significant when the molecules are close together on average. II. Attractive intermolecular forces are important when the molecules are fairly close together but not necessarily touching. III. Attractive forces are ineffective when the molecules are for apart. Select correct statements.
View More Questions
Top MET Nuclear physics Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions