Respuesta :
Answer:
[tex]K.E =0.081eV[/tex]
[tex]\vec{\Omega}=(0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]
Explanation:
Given:
Velocity vector [tex]\vec{v}=(2132\hat{i}+3300\hat{j}+154\hat{k})m/s[/tex]
the mass of neutron, m = 1.67 × 10⁻²⁷ kg
Now,
the kinetic energy (K.E) is given as:
[tex]K.E =\frac{1}{2}mv^2[/tex]
[tex]\vec{v}^2 = \vec{v}.\vec{v}[/tex]
or
[tex]\vec{v}^2 = (2132\hat{i}+3300\hat{j}+154\hat{k}).(2132\hat{i}+3300\hat{j}+154\hat{k})[/tex]
or
[tex]\vec{v}^2 =15.45\times 10^6 m^2/s^2[/tex]
substituting the values in the K.E equation
[tex]K.E =\frac{1}{2}1.67\times 10^{-27}kg\times 15.45\times 10^6 m^2/s^2[/tex]
or
[tex]K.E =1.2989\times 10^{-20}J[/tex]
also
1J = 6.242 × 10¹⁸ eV
thus,
[tex]K.E =1.2989\times 10^{-20}\times 6.242\times 10^{18}[/tex]
[tex]K.E =0.081eV[/tex]
Now, the direction vector [tex]\vec{\Omega}[/tex]
[tex]\vec{\Omega}=\frac{\vec{v}}{\left | \vec{v} \right |}[/tex]
or
[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{\left |\sqrt{((2132\hat{i})^2+(3300\hat{j})^2+(154\hat{k}))^2} \right |}[/tex]
or
[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{3930.65}[/tex]
or
[tex]\vec{\Omega}=0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]
