${}_{n}X_m$ emitted one $\alpha and\mathrm{2 }\beta$ particles, then it will become: 

1. ${}_{n}X_{m-4}$

2. ${}_{n-1}X_{m-1}$

3. ${}_{n}Z_{m-4}$

4. None of these

For the given reaction, the particle \(\mathrm{X}\) is:
\({ }_6^{11} \mathrm{C}\rightarrow { }_5^{11}\mathrm{B}+\beta^{+}+\mathrm{X}\)
1. neutron
2. anti-neutrino
3. neutrino
4. proton

Which rays contain (+ve) charged particle:
1.  α-rays
2.  β-rays
3.  γ-rays
4.  X-rays

$\mathrm{If\; X}(n,$ $\alpha$ $)\; converts\; into$ $$ ${}_3{}^{7}Li$, then X will be: 

1. ${}_5{}^{10}B$

2.  ${}_5{}^{9}B$

3. ${}_4{}^{11}Be$

4. ${}_2{}^{4}He$

If a \({}_{a}^{b}\mathrm{X}\) species emits firstly a positron, then two \(\alpha\) and two \(\beta\) and at last one \(\alpha\) is also emitted and finally convert into stable \({}_{d}^{c}\mathrm{Y}\) species, so the correct relation will be:
1. \(c = b-12, d = a-5\)
2. \(a = c-8, d = b-1\)
3. \(a = c-6, d = b-0\)
4. \(a = c-4, a = b-2\)

If a deuteron is bombarded on a ₈O¹⁶ nucleus then an α-particle is emitted. The product nucleus is:

1. ₇N¹³

2. ₅B¹⁰

3. ₄Be⁹

4. ₇N¹⁴