Physics, 02.09.2019 16:10 lilfofo9

Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutron turns into a proton, which is retained in the nucleus, and emits radiation in the form of an electron. what is the product of such radioactive decay of phosphorus-32?

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Answer from: jonathanmagana112002

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

Answer from: Quest

heats surface water up

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.