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Tpy6a [65]
3 years ago
15

In a photoelectric experiment a student uses a light source whose frequency is greater than that needed to eject electrons from

a certain metal. However, after continuously shining the light on the same area of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency of the light is held constant. How would you account for this behaviour?
Chemistry
1 answer:
ankoles [38]3 years ago
8 0

Answer:

The metal surface becomes more positive as electrons are lost from it.

Explanation:

Let us note that photoelectric effect refers to a phenomenon in which electrons are ejected from a clean metal surface irradiated with light of appropriate frequency. This photon must possess a frequency above the threshold frequency of the metal and its energy must exceed the work function of the metal. When these conditions are met, electrons are emitted from a clean metal surface, having a constant kinetic energy as long as the frequency of the incident photon remains constant.

However, as photoelectric effect progresses and electrons are lost from the metal surface, the metal surface becomes more positive. The more positive the surface, the greater the attraction of the positive surface for the emitted electrons. This reduces the kinetic energy of the emitted photons even though the frequency of incident photons is held constant.

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An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?
aleksley [76]

Answer:- Formula of the hydrate is Fe_2(SO_4)_3.5H_2O and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})

= 0.204molFe_2(SO_4)_3

18.4gH_2O(\frac{1mol}{18.02g})

= 1.02molH_2O

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

Fe_2(SO_4)_3=\frac{0.204}{0.204}  = 1

H_2O=\frac{1.02}{0.204} = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is Fe_2(SO_4)_3.5H_2O and the name of the hydrate is Iron(III)sulfate pentahydrate.


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cestrela7 [59]

Answer: migrate south

Explanation:

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