A) \[_{5}B{{e}^{11}}\]
B) \[_{5}B{{e}^{10}}\]
C) \[_{5}B{{e}^{9}}\]
D) \[_{5}{{C}^{12}}\]
Correct Answer: B
Solution :
\[\text{mg sin 60 = Bil cos 6}{{\text{0}}^{\text{o}}}\] means the nucleus splits into \[B=\frac{mg}{il}\tan {{60}^{\circ }}\]particle and neutrons i.e. \[=\frac{0.01\times 10}{173\times 0.1}\times \sqrt{3}\] So A + 1 = 7 + 4 \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{4\pi a}\left[ \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right]\] A = 10 and Z + 0 = 3 + 2 Z = 5 Hence, the nucleus of element Y is boron i.e. \[B=3{{B}_{1}}=\frac{3{{\mu }_{0}}I}{4\pi a}\left[ \sin {{\theta }_{1}}+\sin {{\theta }_{2}} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
