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3 years ago

What happens to electrons in the photoelectric effect?

2 answers:

5 0

The photoelectric effect occurs when light shines on a metal. ... Light of any frequency will cause electrons to be emitted.

timofeeve [1]3 years ago

4 0

Answer:

Explanation:

Photoelectric effect was discovered in 1887 by Heinrich Rudolf Hertz. It refers to the ejection or emission of electrons when light hits a material (generally metal). Emitted electrons are called photo-electrons but the intensity of the light determine the kinetic energy of emitted electron.

Intensity of light is directly proportional to the kinetic energy emitted by an electron, that means more intensity the light have more will be the kinetic energy emitted by an electron.

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How many grams of NaNO3 in 165 grams of H2O at 20°C would fully dissolve?

Fantom [35]

Answer: birchbox

Explanation:

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3 years ago

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.0 \times {10^{15}}\;{\text{Hz}}$ is $\boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}$

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\mathbf{E=h\nu }}$ ......(1)

Here,

${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays is $1.0 \times {10^{15}}\;{\text{Hz}}$ or $1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}$

The value of Planck’s constant is $6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}}$ .

Substitute these values in equation (1)

$\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}}$ .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

7 0

3 years ago

Read 2 more answers

From the lists of available reagents select the one(s) you would use to in a preparation of acetophenone (phenyl methyl ketone)

VARVARA [1.3K]

Answer:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

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3 years ago

A lawnmower runs on gasoline.what type of energy transformation takes place in its engine?

Eva8 [605]

The gas is ignited (I think) and combustion happens where the gasoline turns into gas (the state of being) and expands, pushing something and making the blades turn so

from stationary to explosive so potentioal to kenetic

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3 years ago

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Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s