Respuesta :
The initial kinetic energy of the proton is given by:
[tex]K= \frac{1}{2} mv^2[/tex]
where [tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass and [tex]v=3.4\cdot 10^7 m/s[/tex] is the initial speed.
As the proton approaches the nucleus, it decelerates because of the repulsive electric field and its kinetic energy converts into electric potential energy. The proton will stop at a distance r from the center of the nucleus, and its potential energy at this distance will be:
[tex]U=k_e \frac{(80e)(e)}{r} [/tex]
where [tex]k_e = 8.99\cdot 10^9 N m^2 C^{-2}[/tex], 80e is the charge of the nucleus of mercury (which contains 80 protons), and [tex]e=1.6\cdot 10^{-19}C[/tex] is the proton charge.
For the conservation of energy,
[tex]K=U[/tex]
Rewriting it, we find
[tex]r=2k_e \frac{(80e)(e)}{mv^2}=3.4 \cdot 10^{-15} m =34 fm [/tex]
This is not the final answer, however, because this is the distance reached by the proton with respect to the center of the nucleus. So, to find the distance from the surface, we should subtract the radius of the nucleus, which is half the diameter: 14/2=7 fm. So
[tex]d=r-r_0=34 fm-7 fm=27 fm =2.7 \cdot 10^{-15 } m[/tex]
[tex]K= \frac{1}{2} mv^2[/tex]
where [tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass and [tex]v=3.4\cdot 10^7 m/s[/tex] is the initial speed.
As the proton approaches the nucleus, it decelerates because of the repulsive electric field and its kinetic energy converts into electric potential energy. The proton will stop at a distance r from the center of the nucleus, and its potential energy at this distance will be:
[tex]U=k_e \frac{(80e)(e)}{r} [/tex]
where [tex]k_e = 8.99\cdot 10^9 N m^2 C^{-2}[/tex], 80e is the charge of the nucleus of mercury (which contains 80 protons), and [tex]e=1.6\cdot 10^{-19}C[/tex] is the proton charge.
For the conservation of energy,
[tex]K=U[/tex]
Rewriting it, we find
[tex]r=2k_e \frac{(80e)(e)}{mv^2}=3.4 \cdot 10^{-15} m =34 fm [/tex]
This is not the final answer, however, because this is the distance reached by the proton with respect to the center of the nucleus. So, to find the distance from the surface, we should subtract the radius of the nucleus, which is half the diameter: 14/2=7 fm. So
[tex]d=r-r_0=34 fm-7 fm=27 fm =2.7 \cdot 10^{-15 } m[/tex]
Answer:
[tex]27fm[/tex]
Explanation:
Kinetic Energy of proton
[tex]Kinetic Energy (K)=\frac{1}{2}mV^{2}[/tex]
[tex]m=Mass of proton[/tex]
[tex]V=Velocity of proton[/tex]
[tex]m=1.67\times 10^{-27} kg[/tex]
[tex]V=3.4\times 10^{7}ms^{-1}[/tex]
[tex]K=\frac{1}{2}\times 1.67\times 10^{-27}kg\times \left (3.4\times 10^{7}ms^{-1} \right )^2[/tex]
[tex]K=\frac{19.305}{2}\times 10^{-13}J[/tex]
[tex]K=9.65\times 10^{-13}J[/tex]
For conservation of energy;
[tex]Kinetic Energy=Potential energy[/tex]
[tex]K= U[/tex]
So,
[tex]U= 9.65\times 10^{-13}J[/tex]
Here,
[tex]U=Potential Energy[/tex]
[tex]U=k_{e}\frac{q_{1}q_{2}}{r}[/tex]
Here,
[tex]k_{e}=Coulomb's law constant[/tex]
[tex]k_{e}=8.99\times 10^{9}Nm^{2}C^{-2}[/tex]
[tex]q_{1}=80e[/tex]
[tex]q_{2}=e[/tex]
[tex]e=1.602\times 10^{-19}C[/tex]
[tex]r=The distance that proton will stop from the center of the nucleus[/tex]
[tex]U=k_{e}\frac{80e\times e}{r}[/tex]
[tex]9.65\times 10^{-13}J=8.99\times 10^{9}Nm^{2}C^{-2}\frac{80\times1.6\times 10^{-19}\times1.6\times 10^{-19} }{r}[/tex]
[tex]r=8.99\times 10^{9}Nm^{2}C^{-2}\frac{80\times1.6\times 10^{-19}\times1.6\times 10^{-19} }{9.65\times 10^{-13}J}[/tex]
[tex]34fm[/tex]
[tex]r_{0}=Radius of the atom[/tex]
[tex]Radius\left ( r_{0} \right )=\frac{diameter\left ( d \right )}{2}[/tex]
[tex]Diameter of the nucleus of mercury atom=14fm[/tex]
[tex]Radius of atom =\frac{14fm}{2}[/tex]
[tex]r=7fm[/tex]
[tex]d=r-r_{0}[/tex]
[tex]d=34fm-7fm[/tex]
[tex]d=27fm[/tex]
Further Explanation:
When a proton approaches a nucleus, it decelerates. Because the repulsive electric field and its kinetic energy converts into electric potential energy.
Then due to this, the proton will stop at a distance “r” from the center of the nucleus.
To find the distance from the surface where the proton hits, we have to subtract the radius of the nucleus.
Learn more:
1. Kinetic energy https://brainly.com/question/1621817 (answer by skyp)
2. Potential energy https://brainly.com/question/12489105 (answer by nitrotype2000)
3. Conservation of energy https://brainly.com/question/11911812 (answer by hrishisup)
Keywords:
Kinetic energy, potential energy, conservation of energy.
