Respuesta :
Answer:
The kinetic energy and speed of the two helium atoms are [tex]73.6\times10^{-16}\ J[/tex] and [tex]14.88\times10^{5}\ m/s[/tex].
Explanation:
Given that,
Mass of beryllium = 8.00 u
Mass of helium = 4.00 u
Energy = 92.2 keV
We need to calculate the kinetic energy
The kinetic energy of the helium atom is the half of the total kinetic energy released.
[tex]K.E_{h}=\dfrac{E_{total}}{2}[/tex]
[tex]K.E_{h}=\dfrac{92.0}{2}[/tex]
[tex]K.E_{h}=46 K.eV[/tex]
Therefore, the kinetic energy of the each helium atom is
[tex]K.E_{h}=\dfrac{1}{2}mv^2[/tex]
[tex]K.E_{h}=\dfrac{92.0\times10^3\times1.6\times10^{-19}}{2}[/tex]
[tex]K.E_{h}=73.6\times10^{-16}\ J[/tex]
We need to calculate the speed
Using formula of speed
[tex]\dfrac{1}{2}\times mv^2=73.6\times10^{-16}\ J[/tex]
[tex]v^2=2\times\dfrac{73.6\times10^{-16}}{m}[/tex]
[tex]v^2=2\times\dfrac{73.6\times10^{-16}}{4.00\times1.66\times10^{-27}}[/tex]
[tex]v^2=221.68\times10^{10}[/tex]
[tex]v=14.88\times10^{5}\ m/s[/tex]
Hence, The kinetic energy and speed of the two helium atoms are [tex]73.6\times10^{-16}\ J[/tex] and [tex]14.88\times10^{5}\ m/s[/tex].
