What is the respective number of $\alpha$ and $\beta$-particles emitted in the following radioactive decay?

$X_{90}^{{}_{200}}\to {}_{}Y_{80}^{168}$ 

1. 6 and 8

3. 6 and 6

3. 8 and 8

4. 8 and 6   

A free neutron decays into a proton, an electron and:

1. A beta particle.

2. An alpha particle.

3. An antineutrino.

4. A neutrino.

A nucleus ${}_n{}^{m}X$ emits one $\alpha$ and two $\beta -$particles. The resulting nucleus is

1. ${}_n{}^{m-4}X$ 

2. ${}_{n-2}{}^{m-4}X$   

3. ${}_{n-4}{}^{m-4}X$ 

4. None of these

In one \(\alpha \text-\)$$ and \(2\beta \text-\)emissions:

A nuclear decay is expressed as:
\(_{6}^{11}\mathrm{C}\rightarrow _{5}^{11}\mathrm{B}+\beta^{+}+\mathrm{X}\)
Then the unknown particle \(X\) is:
1. neutron 
2. antineutrino
3. proton 
4. neutrino

When a deuterium is bombarded on ${}_{8}O_{16}$ nucleus, an $\alpha$-particle is emitted, then the product nucleus is:

1. ${}_{7}N_{13}$ 

2. ${}_{5}B_{10}$

3. ${}_{4}B{e}_9$ 

4. ${}_{7}N_{14}$

A nuclear reaction given by \({ }_{Z}^{A} \mathrm{~X} \rightarrow{ }_{Z+1}^{A} \mathrm{Y}+e^{-}+\bar{v}\) represents:
1. fusion
2. fission
3. \(\beta^{-} \text-\)decay
4. \(\gamma \)-decay

The nucleus ${}_6{}^{12}C$ absorbs an energetic neutron and emits a beta particle $\left({\beta }^{-}\right)$. The resulting nucleus is 

1. ${}_7{}^{14}C$

2. ${}_7{}^{13}N$ 

3. ${}_5{}^{13}B$ 

4. ${}_6{}^{13}C$

A radioactive nucleus undergoes a series of decay according to the scheme 

$A\stackrel{\alpha }{\to }{A}_1\stackrel{\beta }{\to }{A}_2\stackrel{\alpha }{\to }{A}_3\stackrel{\gamma }{\to }{A}_4$

If the mass number and atomic number of A are 180 and 72 respectively, then what are these number for $A_4$?

1. 172 and 69 

2. 174 and 70

3. 176 and 69

4. 176 and 70

An element \(A\) decays into element \(C\) by a two-step process
\(A\rightarrow B+_{2}^{4}\mathrm{He}\)
\(B\rightarrow C+2e^{-}\)
Then: