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2 years ago

Which of the following was most clearly, even uniquely perhaps, revealed by the discovery of the photoelectric effect?

Chemistry

2 answers:

nika2105 [10]2 years ago

7 0

Answer : Option 3) Wave/Particle duality.

Explanation : The experiment on discovery of photoelectric effect revealed about the photoelectrons of light that can behave as particle or waves.

The photoelectric effect is observed when the emission of electrons or other free carriers occurs on shining a light radiation on a material. The electrons emitted from this can be called photo electrons. These photoelectrons may behave as wave or particle in duality which holds that light and matter exhibit properties of both waves and of particles.

kondaur [170]2 years ago

5 0

The discovery of the photoelectric effect reveals $\boxed{{\text{3}}{\text{.}}\;{\text{wave - particle duality}}}$ .

Further explanation:

Photoelectric effect is the phenomenon in which metals having certain minimum frequency emits electron when exposed to electromagnetic radiation. The electrons emitted are called as photoelectrons.

The electrons are emitted as a result of absorption of energy from electromagnetic radiation of high frequency. Electromagnetic radiation such as ultraviolet light excites electrons to higher energy level called as an excited state of electrons.

Some features of photoelectric effects are written as follows:

1. The energy of photon is proportional to frequency of light.

2. A certain minimum frequency called as threshold frequency is needed by metal to emit electrons below which no electron are emitted. It is denoted by ${\nu }}_{\text{o}}}$ .

3. Kinetic energy of electron is proportional to frequency of electromagnetic radiation.

4. Number of electrons emitted per second is proportional to intensity of light.

Wave-particle duality is the phenomenon in which photon exhibit properties of both particle and wave.

According to photoelectric effect, light can exhibit both particle as well as wave nature. Each photon carries definite amount of energy that is directly proportional to frequency and behaves as a particle.

On the other hand, when light hits metal surface having certain minimum frequency it causes emission of electrons. The minimum amount of energy and frequency translates into a maximum wavelength. Therefore light propagates and interacts with metal surface to eject electron as it is a wave.

Hence the phenomenon of wave-particle duality is best revealed by discovery photoelectric effect.

Learn more:

1. What is the charge associated with each molecule of HCl brainly.com/question/5288267

2. Write the chemical equation brainly.com/question/8926688

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Photoelectric effect

Keywords: Photon, photoelectric effect, energy, metals, light, particle, electromagnetic radiation, frequency, threshold frequency, kinetic energy and wave-particle duality.

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The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),

choli [55]

Answer:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

B. 100.00° C

C. 101.02° C

D. 100.51° C

Which of the following solutions will have the lowest freezing point?

A. 1.0 mol/kg sucrose (C12H22O11)

B. 1.0 mol/kg lithium chloride (LiCl)

C. 1.0 mol/kg sodium phosphide (Na3P)

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

B. magnesium sulfate (MgSO4)

C. potassium bromide (KBr)

D. All will be equally effective.

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

A. 98.7 kPa

B. 96.3 kPa

C. 101.3 kPa

D. 100.2 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

A. 101.2°C

B. 105.9°C

C. 102.7°C

D. 108.1°C

Explanation:

The following are the answers to the different questions:

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C

5 0

2 years ago

A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

ioda

In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = $\frac{PV}{nT}$ = constant
$\frac{P1 V1}{n T}$ = $\frac{P2 V2}{n T}$
==> P1 * V1 = P2 * V2
V2 = $\frac{P1 V1}{P2}$ = $\frac{445 0.016}{1225}$
V2 = 0.00581 m3 = 5.81 L

7 0

2 years ago

To determine the concentration of ethanol in cognac a 5.00 mL sample of the cognac is diluted to 0.500 L. Analysis of the dilute

julia-pushkina [17]

Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molar concentration of undiluted cognac = ?

$M_2$ = molar concentration of diluted cognac = 0.0844 M

$V_1$ = volume of undiluted cognac = 5.00 mL = 0.005 L

$V_2$ = volume of diluted cognac = 0.500 L

Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.

$M_1\times 0.005L=0.0844M\times 0.500L$

$M_1=8.44M$

Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M

4 0

2 years ago

A 0.1025-g sample of copper metal is dissolved in 35 ml of concentrated hno3 to form cu2 ions and then water is added to make a

GalinKa [24]

Molarity is defined as number of moles of solute in 1 L of solution.

Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.

The balanced reaction will be as follows:

Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O

From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:

n=\frac{m}{M}

molar mass of Cu is 63.55 g/mol thus,

n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol

Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:

M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M

Thus, molarity of Cu^{2+} is 0.008 M.

7 0

2 years ago

During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S

Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of $^{235}UF_6=349.034348g/mol$

Molar mass of $^{238}UF_6=352.041206g/mol$

By taking their ratio, we get:

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}$

$\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}$

From the above relation, it is clear that rate of effusion of $^{235}UF_6$ is faster than $^{238}UF_6$

Difference in the rate of both the gases, $Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816$

To calculate the percentage increase in the rate, we use the equation:

$\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100$

Putting values in above equation, we get:

$\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%$

The above calculations proves that the rate of diffusion of $^{235}UF_6$ is 0.4 % faster than the rate of diffusion of $^{238}UF_6$

3 0

2 years ago