Boron has two isotopes ${}_{5}B^{10}$ and ${}_{5}B^{11}$. If atomic weight of Boron is 10.81 then ratio of ${}_{5}B^{10}$ to ${}_{5}B^{11}$in nature will be: 

1. 15 : 16 

2. 19: 81

3. 81 : 19 

4. 20: 53

${}_{n}X_m$ 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. ${}_2{}^{4}He$