Answer:
[Ne]3s2
Explanation:
ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.
considerando la configuración electrónica más externa de cada una de las especies mostradas;
para la primera configuración, ns2 np6 corresponde a un gas noble.
para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.
para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.
para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6
The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980
Answer:
The second answer choice
Explanation:
X seems to list attributes for opaque objects, and y lists attributes of Transparent objects.
Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number 
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as 
, 
and 
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
I believe that it is a chemical change.
