Question

Complete the following nuclear equations:
 \[{}_7^{14}N + {}_2^4He\, \to {}_8^{17}O\, + .....\] \[{}_4^9Be + {}_2^4He\, \to {}_6^{12}C\, + .....\] \[{}_4^9Be(p,\alpha )....\]
 \[{}_{15}^{30}P \to {}_{14}^{30}Si....\] \[{}_1^3H \to {}_2^4He....\] \[{}_{20}^{43}Ca(\alpha ,...) \to {}_{21}^{46}Sc....\]
A. \[(a){}_1^1H\,(b)\,{}_0^1n\,\,(c)\,{}_3^6Li\,(d){}_{ + 1}^0e(e){}_{ - 1}^0e\,(f)p\,(protone)\]
B. \[(a){}_1^1H\,(b)\,{}_0^1n\,\,(c)\,{}_{32}^6Li\,(d){}_{ - 1}^0e(e){}_{ - 1}^0e\,(f)p\,(protone)\]
C. \[(a){}_1^1H\,(b)\,{}_0^1n\,\,(c)\,{}_{ - 3}^6Li\,(d){}_{ - 1}^0e(e){}_{ - 1}^0e\,(f)p\,(protone)\]
D. \[(a){}_1^1H\,(b)\,{}_0^1n\,\,(c)\,{}_3^6Li\,(d){}_{ - 1}^0e(e){}_{ + 1}^0e\,(f)p\,(protone)\]

Answer 1

Hint: Radioactive decay is the loss of elementary particles from an unstable nucleus, ultimately changing the unstable element into another more stable element. There are various types of decay. Each type of decay emits a specific particle that changes the type of product produced. The number of protons and neutrons found in the daughter nuclei are determined by the type of decay or emission that the original element goes through.

Complete step by step answer:
 Radioactive decay is of three types, alpha decay, beta decay, gamma decay. In alpha decay, one particle with atomic number 2 and atomic mass 4 gets removed from the nuclei. In beta decay, two kinds of the process are possible one is positive beta decay and another one is negative beta decay. In the case of gamma decay, no change of nuclei happens.
In a radioactive reaction, the mass of atoms on the left-hand side should be the same as the right-hand side. Therefore, the given reactions are,
 \[{}_7^{14}N + {}_2^4He\, \to {}_8^{17}O\, + {}_1^1H\]
In these reactions emission of protons takes place, by which the mass in L.H.S is equal to R.H.S.
 \[{}_4^9Be + {}_2^4He\, \to {}_6^{12}C\, + {}_0^1n\]
In these reactions emission of neutrons takes place, by which the mass in L.H.S is equal to R.H.S.
 \[{}_4^9Be(p,\alpha ){}_3^6Li\]
This is a representation of a radioactive reaction; the reaction is as follows.
 \[{}_4^9Be + {}_1^1H \to {}_3^6Li + {}_2^4He\]
 \[{}_{15}^{30}P \to {}_{14}^{30}Si + \,{}_{ + 1}^0e\]
This is a radioactive reaction of beta decay; here positive beta decay has happened. by which the mass in L.H.S is equal to R.H.S.
  \[{}_1^3H \to {}_2^4He + {}_{ - 1}^0e\]
 \[{}_{15}^{30}P \to {}_{14}^{30}Si + \,{}_{ + 1}^0e\]
This is a radioactive reaction of beta decay; here negative beta decay happens. by which the mass in L.H.S is equal to R.H.S.
 \[{}_{20}^{43}Ca(\alpha ,...) \to {}_{21}^{46}Sc. + {}_1^1H(protone)\]
In these reactions emission of protons takes place, by which the mass in L.H.S is equal to R.H.S.

So, the correct answer is A.

Note:
Alpha particles are produced during standard radioactive decay of radioactive elements. These are emitted in the form of radiation, having an average kinetic energy of 5MeV and velocity \[5\% \] as compared to the speed of light. They are also produced in accelerators of high energy particles. They are highly ionized and have low penetration power.