Assertion : For maximum stability N/P ratio must be equal to 1
Reason : Loss of `alpha - beta -` particles has no role in N/P ratio
Assertion : For maximum stability N/P ratio must be equal to 1
Reason : Loss of `alpha - beta -` particles has no role in N/P ratio
Reason : Loss of `alpha - beta -` particles has no role in N/P ratio
A
If both assertion and reason are true and the reason is the correct explanation of the assertion
B
If both assertion and reason are true but reason is not the correct explanantion of the assertion
C
If assertion is true but reason is false
D
If the assertion and reason both are false
Text Solution
AI Generated Solution
The correct Answer is:
To solve the question, we need to analyze both the assertion and the reason provided.
### Step 1: Analyze the Assertion
The assertion states that "For maximum stability, the N/P (neutron-to-proton) ratio must be equal to 1."
- **Explanation**: The neutron-to-proton ratio is crucial for the stability of a nucleus. For lighter elements, a ratio close to 1 is often observed for maximum stability. As elements become heavier, the ratio tends to increase (more neutrons than protons) to counteract the repulsive forces between the positively charged protons. Therefore, while the assertion is somewhat simplified, it is generally correct for lighter elements.
### Step 2: Analyze the Reason
The reason states that "Loss of alpha and beta particles has no role in N/P ratio."
- **Explanation**: This statement is incorrect. The loss of alpha particles (which consist of 2 protons and 2 neutrons) will decrease both the number of neutrons and protons, thus affecting the N/P ratio. Similarly, the emission of beta particles involves the conversion of a neutron into a proton (or vice versa), which directly alters the N/P ratio. Therefore, the reason provided does not support the assertion.
### Step 3: Conclusion
Based on the analysis:
- The assertion is correct, as a neutron-to-proton ratio of 1 is ideal for stability in lighter nuclei.
- The reason is incorrect because the loss of alpha and beta particles does indeed affect the N/P ratio.
### Final Answer
- **Assertion**: True
- **Reason**: False
To solve the question, we need to analyze both the assertion and the reason provided.
### Step 1: Analyze the Assertion
The assertion states that "For maximum stability, the N/P (neutron-to-proton) ratio must be equal to 1."
- **Explanation**: The neutron-to-proton ratio is crucial for the stability of a nucleus. For lighter elements, a ratio close to 1 is often observed for maximum stability. As elements become heavier, the ratio tends to increase (more neutrons than protons) to counteract the repulsive forces between the positively charged protons. Therefore, while the assertion is somewhat simplified, it is generally correct for lighter elements.
### Step 2: Analyze the Reason
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Similar Questions
Explore conceptually related problems
In alpha- decay , n//p ratio :
Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio (n//p) . For stable nuclei n//p ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having n//p ratio either very high or low undergo nuclear transformation. When n//p ratio is higher than required for stability, the nuclei have the tendency to emit beta -rays. while when n//p ratio is lower than required for stability, the nuclei either emits alpha -particles or a positron or capture K -electron. beta -particle is emitted in radioactivity by
In beta- decay n//p ratio :
Assertion : Alpha particles produce more intense ionisation than beta particles. Reason : Alpha particles are positively charged .
The stabiliy of nuclies has been explained in terms of binding energy. Higher is binding energy, more is stabliliy to nucleus. The binding energy is written as B.E = 931.478 xx Delta m' MeV m wehre Delta m' is mass decayed in amu. If B.E.//"nucleon" lies below the belt of stablilty, the nucleus undergoes alpha- emission in order to lower down the energy level of nucleus but its n//p ratio increases. To lower down the energy level of nucleus but is n//p ratio increases. To lower down level of nucleus loses beta- particles and if stability is not gained, gamma- emission is noticed. A radioactive element on losing on alpha- particles shows a loss in its mass number by 4 units and atomic number by 2 units whereas beta- emission leads to a gain in at.no by one units and mass no. remains same. An element having n//p ratio greater than 1 will show:
Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio (n//p) . For stable nuclei n//p ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having n//p ratio either very high or low undergo nuclear transformation. When n//p ratio is higher than required for stability, the nuclei have the tendency to emit beta -rays. while when n//p ratio is lower than required for stability, the nuclei either emits alpha -particles or a positron or capture K -electron. For reaction ._(92)M^(238) rarr ._(y)N^(x) + 2 ._(2)He^(4), ._(y)N^(x) rarr ._(B)L^(A) + 2 ._(-1)e^(0) The number of neutrons in the element L is
A proton and an alpha - particle, having kinetic energies K_P and K_a respectively enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of alpha- particle is 2:1. The ratio of K_p : K_a is :
A : During radioactive disintegration an alpha -particle and a beta -particle do not emit simultaneously from any nucleus. R : An alpha -particle emits from a nucleus when the N/Z ratio is less than the stability range (where N=number of neutrons and Z=number of protons in a nucleus. )
Assertion (A) : beta- particles are deflected more than alpha- particles in a given electric field. Reason (R) : Charge on alpha- particles is larger than on beta- particles.
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