Chemistry-mole calculations

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

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Sugar costs $0.98/kg. 1 metric ton=1000 kg. How many metric tons of sugar can

EastWind [94]

Answer:

0.35714 Metric ton

Explanation:

350 / 0.98 = 357.1 ----------A

to get the answer on metric ton divide (a) by 1000

357.14/1000= 0.357

7 0

3 years ago

Which of the following is not a property of a buffer solution? a) It is a mixture of a weak acid and its conjugate base. b) Resi

m_a_m_a [10]

Answer:

e) pH is independent of concentration.

Explanation:

a) It is a mixture of a weak acid and its conjugate base. TRUE. A buffer is defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

b) Resists pH changes because it reacts with added acid or base. TRUE. Thermodynamically, the reaction of added acid or base is faster with the buffer mixture than with H⁺ or OH⁻ ions of the solutions.

c) The maximum buffer capacity is at pH = pKa. TRUE. The buffer capacity is pka±1. For this, buffer capacity is maximum in pka.

d) pH is dependent on the solution ionic strength and temperature. TRUE. Ionic strength and temperature are factors that influence concentrations of ions in solutions as the H⁺ ion that is the responsible

e) pH is independent of concentration. FALSE. pH in a buffer depends completely of concentrations of the acid and its conjugate base or vice versa.

I hope it helps!

8 0

3 years ago

How many particles are in 47.7 g of Magnesium? (Round the average

Lynna [10]

Answer:

1.18 × 10²⁴ particles Mg

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

47.7 g Mg

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Mg - 24.31 g/mol

Step 3: Convert

$47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} )$ = 1.18161 × 10²⁴ particles Mg

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg

3 0

2 years ago

The decomposition of mercury (ii) oxide at high temperature, is it an endothermic or exothermic process? Write a chemical reacti

gulaghasi [49]

Answer:

that will be endothermic reaction...... as oxides of mercury decomposes and break s into simpler elements by absorbing energy

Explanation:

hope it helped u buddy

6 0

3 years ago

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