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:
B. 8
Explanation:
because I learned that in 6th
Answer:
a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.
b): not necessarily due to London Dispersion Forces.
Explanation:
There are three major types of intermolecular interaction:
- Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
- Dipole-dipole interactions between all molecules.
- London dispersion forces between all molecules.
The melting point of a substance is a result of all three forces, combined.
Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.
For example, 
(water) molecules are capable of hydrogen bonding. The melting point of at 
is around 
. That's considerably high when compared to other three-atom molecules.
In comparison, the higher alkane hexadecane (
, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around 
above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.
Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around 
than that of HCl.
