Which part of an atom is mostly empty space?

Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutro

dimulka [17.4K]

Answer:

The product of the decay its Sulfur-32

Explanation:

Phosphorus-32 ( lets write it $_{15}^{32}P$ , where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:

$_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}$

where $e^-$ its the electron, and $\bar{\nu_e}$ the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).

So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:

$_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}$ .

$_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}$ .

The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its

$_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}$ .

and the product of such decay its Sulfur-32

5 0

4 years ago

If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

Arada [10]

Answer:

8.93*10^13 N.

Explanation:

Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:

$F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }$

where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.
Replacing by the values, we get:

$F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N$

Fg = 8.93*10^13 N.

6 0

3 years ago

This question is typical on some driver’s license exams: A car moving at 45 km/h skids 15 m with locked brakes. How far will the

sergiy2304 [10]

from kinematics equation if we know that final speed is ZERO and initial speed is given that due to constant deceleration the object will stop in some distance "d" and this distance can be calculated by kinematics

$v_f^2 - v_i^2 = 2 a d$