Chemistry, 18.10.2019 11:30 mayaleila

What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.3×1015hz?

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The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.3\times{10^{15}}\;{\text{Hz}}$ is $\boxed{8.614\times10^{-19}\text{ J}}$ .

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 the photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to the 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:

${\text{E}}={\text{hv}}$ …… (1)

Here,

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

is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays $1.3\times{10^{15}}\;{\text{Hz}}$ is or $1.3\times{10^{15}}\;{{\text{s}}^{-1}}$

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

Substitute these values in equation (1)

$\begin{gathered}{\text{E}}=\left({6.626\times{{10}^{-34}}\;{\text{J}}\cdot{\text{s}}}\right)\left({1.3\times{{10}^{15}}\;{{\text{s}}^{-1}}}\right)\\= 8.614\times{10^{-19}}\;{\text{J}}\\\end{gathered}$

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 ${\mathbf{8}}{\mathbf{.614\times1}}{{\mathbf{0}}^{{\mathbf{-19}}}}\;{\mathbf{J}}$

Learn more:

1. Calculate the frequency of n = 6 in the Lyman series:

2. The energy of a photon in light:

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.

Answer from: Quest

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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