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natali 33 [55]
2 years ago
5

A metal foil has a threshold frequency of 5.45 \times 10^{14} \text{ } \text{Hz}5.45×10 14 Hz. Which of the colors of visible l

ight have enough energy to eject electrons from this metal? Select all that are correct.Red
Orange
Yellow
Green
Blue
Indigo
Violet

Chemistry
2 answers:
mash [69]2 years ago
8 0

Explanation:

Below is an attachment containing the solution

nalin [4]2 years ago
7 0

Answer:

Green, Blue, Indigo, Violet

Explanation:

According to Einstein's equation of photoelectric effect, the kinetic energy of emitted photoelectron is the difference between the energy of the incident photon and the work function of the metal. The work function of the metal referee to the minimum energy that must be supplied in order to eject an electron from a metal surface. The energy of the incident photon must exceed the work function of the metal.

When we look at the electromagnetic spectrum, only the selected colours have frequency above the threshold frequency as shown by the image attached below.

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Answer:

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Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)

The half-cell reactions are:

Oxidation half reaction (anode):  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Pb^{2+}+2e^-\rightarrow Pb

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

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Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

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See the image below-⬇⬇-

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Answer:

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