Home
Class 12
CHEMISTRY
.(15)P^(29) has n//p ratio too low for s...

`._(15)P^(29)` has `n//p` ratio too low for stability. Its stability can be increased by
a. Positron emission b. Beta-decay
c. Alpha-decay d. Electron capture

A

a. Positron emission

B

b. Beta-decay

C

c. Alpha-decay

D

d. Electron capture

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the stability of the isotope \( _{15}^{29}P \) (Phosphorus-29) and how to increase its stability, we can follow these steps: ### Step 1: Understand the Neutron to Proton Ratio The stability of a nucleus is often determined by the ratio of neutrons (n) to protons (p). For stable nuclei with atomic numbers less than 20, this ratio should ideally be close to 1. ### Step 2: Identify the Current Neutron to Proton Ratio For phosphorus-29: - Atomic number (Z) = 15 (which is the number of protons) - Mass number (A) = 29 (which is the total number of protons and neutrons) To find the number of neutrons (N): \[ N = A - Z = 29 - 15 = 14 \] Now, we can calculate the neutron to proton ratio: \[ \text{n/p ratio} = \frac{N}{Z} = \frac{14}{15} \approx 0.93 \] Since this ratio is less than 1, it indicates that the isotope is unstable. ### Step 3: Determine How to Increase Stability To increase the stability of the nucleus, we need to increase the neutron to proton ratio. This can be achieved by decreasing the number of protons or increasing the number of neutrons. ### Step 4: Analyze the Given Options - **a. Positron emission**: This process involves the emission of a positron (a positive electron), which decreases the number of protons while keeping the number of neutrons the same. This will increase the n/p ratio. - **b. Beta-decay**: This process involves the conversion of a neutron into a proton, which would decrease the n/p ratio, making the nucleus less stable. - **c. Alpha-decay**: This process involves the loss of 2 protons and 2 neutrons. While it decreases the atomic number, it does not necessarily increase the n/p ratio effectively for this case. - **d. Electron capture**: This process involves a proton capturing an electron and turning into a neutron, which increases the number of neutrons but does not help in this specific case since it does not directly decrease the number of protons. ### Step 5: Conclusion The best option to increase the stability of \( _{15}^{29}P \) is **a. Positron emission**, as it decreases the number of protons and thus increases the neutron to proton ratio. ### Final Answer: **a. Positron emission** ---
To solve the question regarding the stability of the isotope \( _{15}^{29}P \) (Phosphorus-29) and how to increase its stability, we can follow these steps: ### Step 1: Understand the Neutron to Proton Ratio The stability of a nucleus is often determined by the ratio of neutrons (n) to protons (p). For stable nuclei with atomic numbers less than 20, this ratio should ideally be close to 1. ### Step 2: Identify the Current Neutron to Proton Ratio For phosphorus-29: - Atomic number (Z) = 15 (which is the number of protons) ...
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Example|18 Videos

  • NUCLEAR CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex6.1 Objective|15 Videos

  • NCERT BASED EXERCISE

    CENGAGE CHEMISTRY ENGLISH|Exercise Nuclear Chemistry (NCERT Exercise)|29 Videos

  • ORGANIC COMPOUNDS WITH FUNCTIONAL GROUP

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Analytical And Descriptive|24 Videos

Similar Questions

Explore conceptually related problems

In alpha- decay , n//p ratio :

In beta- decay n//p ratio :

Reaction intermdiates are short lived species and are highly reactive. They are formed by heterolytic and homolytic bond fission. There are various types of reaction intermediates in which the most important are carbocation , carbanion and free radical. Carbocation is an organic species in which carbon have positive charge and six electrons in its outermost shell. The stability of carbocation can be increased by positive inductive effect, hyperconjugation and delocalisation. If alpha -atom with respect to carbocation has one or more lone pair of electron then lone pair of electron strongly stabilises the carbocation due to octet completion. Species in which carbon have negative charge is called carbanion. Carbanion carries three bond pairs and one lone pair. The stability of carbanion can be increased by negative inductive effect, negative mesomeric effect and delocalisation. Free radical is a species which have seven electrons in its outermost shell. The stability of free radical can be increased by hyperconjugation and delocalisation. The stability order of following free radicals is: C_(6)H_(5)underset(I)CH_(2)overset(*)CH_(2)" "CH_(3)underset(II)CH_(2)overset(*)CH_(2)" "underset(" "III)(C_(6)H_(5))overset(*)CH_(2)" "underset(IV)(CH_(3)^(*)

When nucleus of an electrically neutral atom undergoes a radioactive decay process, it will remain neutral after the decay if the process is (a) An alpha - decay (b) A beta^(o+) -decay (c ) A gamma -decay (d) A K - capture process

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. Unstalbe substance exhibit high radioactivity due to

Reaction intermdiates are short lived species and are highly reactive. They are formed by heterolytic and homolytic bond fission. There are various types of reaction intermediates in which the most important are carbocation , carbanion and free radical. Carbocation is an organic species in which carbon have positive charge and six electrons in its outermost shell. The stability of carbocation can be increased by positive inductive effect, hyperconjugation and delocalisation. If alpha -atom with respect to carbocation has one or more lone pair of electron then lone pair of electron strongly stabilises the carbocation due to octet completion. Species in which carbon have negative charge is called carbanion. Carbanion carries three bond pairs and one lone pair. The stability of carbanion can be increased by negative inductive effect, negative mesomeric effect and delocalisation. Free radical is a species which have seven electrons in its outermost shell. The stability of free radical can be increased by hyperconjugation and delocalisation. The stability order of following carbocations is

._(83)^(214)Bi decays to A by alpha -emission . A then decays to B by beta emission , which further decays to C by another beta emission . Element C decays to D by still another beta emission , and D deacays by alpha -emission to form a stable isotope E. What is element E?

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

During alpha-decay , a nucleus decays by emitting an alpha -particle ( a helium nucleus ._2He^4 ) according to the equation ._Z^AX to ._(Z-2)^(A-4)Y+._2^4He+Q In this process, the energy released Q is shared by the emitted alpha -particle and daughter nucleus in the form of kinetic energy . The energy Q is divided in a definite ratio among the alpha -particle and the daughter nucleus . A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta -decay .This process also involves a release of definite energy . Initially, the beta -decay was represented as ._Z^AX to ._(Z+1)^AY + e^(-)"(electron)"+Q According to this reaction, the energy released during each decay must be divided in definite ratio by the emitted e' ( beta -particle) and the daughter nucleus. While , in alpha decay, it has been found that every emitted alpha -particle has the same sharply defined kinetic energy. It is not so in case of beta -decay . The energy of emitted electrons or positrons is found to vary between zero to a certain maximum value. Wolfgang Pauli first suggested the existence of neutrinoes in 1930. He suggested that during beta -decay, a third particle is also emitted. It shares energy with the emitted beta particles and thus accounts for the energy distribution. During beta^+ decay (positron emission) a proton in the nucleus is converted into a neutron, positron and neutrino. The reaction is correctly represented as

An unstable nuclide with N/P ratio more than that required for stability can attain stability by