Part CWhat is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ?Express your answer in joules to three significant figures.• View Available Hint(s)Ην ΑΣφ?X•10"KE=JSubmitPrevious Answers I RevElectrons are emitted from the surface of a metal when it's exposed tolight. This is called the photoelectric effect. Each metal has a certainthreshold frequency of light, below which nothing happens. Right atthis threshold frequency, an electron is emitted. Above this frequency,the electron is emitted and the extra energy is transferred to theelectron.Here are some data collected on a sample of cesium exposed to various energies of light.Light energy(eV)Electron KE(eV)Electron emitted?3.87noThe equation for this phenomenon is3.88noKE= hv – hvo3.89yeswhere KEis the kinetic energy of the emitted electron,h = 6.63 x 10 34 J.s is Planck's constant, v is the frequency of the3.90yes0.01light, and vy is the threshold frequency of the metal.3.91yes0.02*Note that 1 eV (electron volt) =1.60 × 10 1º J.Also, since E = hv, the equation can also be written asKE=E-6Part Awhere E is the energy of the light and ø is the binding energy of theelectron in the metal.

Kinetic energy of cesium E = hv Where h = plank's constant = 6.626 ×10-34 J s v = frequency of…

Part C What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ? Express your answer in joules to three significant figures. • View Available Hint(s) Ην ΑΣφ ? x" X IXI X•10" КЕ- J Submit Previous Answers

Part C What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ? Express your answer in joules to three significant figures. • View Available Hint(s) Ην ΑΣφ ? x" X IXI X•10" КЕ- J Submit Previous Answers

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Q: Wave length (nanometers) 2.115 E-12 Joules ?

A: Given, energy = 2.115×10-12 J We have to calculate the wave length in nanometers.

Q: The energy of a photon is 1.05 x 10-19 J. What is the wavelength of the corresponding light (nm)?…

Q: (a) Calculate the wavelength of a photon with an energy equivalent to that of an O=O bond ( 120…

A: An energy equivalent to that of an O=O bond ( 120 kcal/mol)

Q: Using the Bohr model, what is the energy in joules of the photon produced when an electron in a He+…

Q: 26. The change in energy for which of the following processes corresponds to the first ionization…

Q: What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency…

A: Given Frequency of Uv rays = 1.1 × 1015 Hz

Q: Identify the correct ranking of effective nuclear charge between N,O, and C? Why is that?

A: The effective nuclear charge is the net positive charge experienced by an electron in a…

Q: Question 5. Below are the orbital diagrams of four different elements. For each diagram indicate the…

Q: Home Work Calculate the energy needed for an electron to move from n; = 3 to ng ? present in the…

Q: Two particles with charges q₁ and q2 are separated by distance d. Arrange these scenarios according…

Q: An electron in a Bohr hydrogen atom is excited fromn 2 to n= 3. What is the energy of this…

Q: If oxygen has energy levels E1=5.26 x 10^-18J and E2 = 6.96 x 10^-18 J what wavelength of…

Q: Question 16 of 16 In Einstein's photoelectric experiment, the kinetic energy (K.E.) of an electron…

A: Given information: Work function of lead = 6.81×10-19 J Wavelength of light = 38.5 nm

Q: Y-ray photons with a wavelength of 2.54 x 10-5 nm.

A: Answer: Relation between energy of photon and wavelength is shown below: E=hCλ Here: h=planck…

Q: What is the wavelength λ of the photon that has been released in Part B? ΔE = −2.04×10−18 J…

A: The relation between energy and wavelength of a radiation is given by Planck-Einstein’s relation,…

Q: What is the energy of H2trapped in a zeolite, which can be modeled by a particle in a cube given nX=…

A: The energy of a 3D cube: Treat the H2 trapped in a three-dimensional cube, whose edge length is…

Q: A neutron (mass 1.7 x 10 g) emitted by radium travels at 3.4 x 107 ± 0.10× 107 (Note: 1 km is 0.62…

A: Given that: Mass of neutron = 1.7 x 10-24 g Velocity of neutron = 3.4 x 107 miles/h Uncertainty in…

Q: SI c = 3.0 x 108 m/s; h = 6.626 x 10 -34 J s; NA = 6.022 x 1023 moles 1 c = À v _AE =hv q=spht x m x…

A: MnO4- (aq)+ H2C2O4(aq)------> Mn+2 (aq) + CO2(g) As per our guidelines we can only solve first…

Q: The photoelectric work function for a certain metal is 221.8 kJ/mol. If the metal absorbs a photon…

Q: Consider an element that reaches its first excited state by absorption of 565.8 nm light. Determine…

A: The objective of the question is to calculate the following. Determine the energy difference between…

Q: Provide brief (1-2 sentence) explanations for the following observations. Your explanations should…

Q: Be sure to answer all parts. How many electrons in an atom can have each of the following quantum…

Q: How many electrons can be assigned the set of quantum numbers n = 4 and l = 2? (A) 2…

A: The given set represent 4d orbital , and maximum 10 electrons can be assigned.

Q: The n value is the principal quantum number, which for H and He is 1. For He, Z = 2 if no electron…

Q: Consider the following energy levels of a hypothetical atom: E4-2.31 x 10-19 J E3-6.11 x 10-19 J…

A: Change in energy of any photon is given as : where , Change in energy h = Planck's…

Q: What is the energy change (in J) and wavelength (in nm) of light emitted when an electron…

Q: The quantum numbers for the electron(s) in the valence shell of thallium (81TI) are: NOTE: the…

Q: An electron in a hydrogen atom travels from energy level n = 2 to energy level n = 3. Would the…

A: According to Bohr's Equation ∆E=-RH(1nf2-1ni2)RH = Rydberg constant

Q: Use Slater's rule to calculate the effective nuclear charge felt by the outermost electron (Zeff) of…

Q: 3 An e- in a hydrogen atam is hea->n=2, what's the wave length in nm?

Q: A Chlorine atom has a first ionization energy of 1251.2 kJ/mol. Light of what frequency is required…

A: Given -> First Ionization energy of chlorine atom => 1251.2 KJ/mole = 1251200 J/mole (1kJ=…

Q: When light is shined on a thin sheet of metal, electrons are released with a kinetic energy of 5.76…

A: Kinetic energy of electrons will be equal to the energy per photon minus the binding energy . Given…

Q: Calculate energy in kJ ?

Concept explainers

Atomic Structure

The basic structure of an atom is defined as the component-level of atomic structure of an atom. Precisely speaking an atom consists of three major subatomic particles which are protons, neutrons, and electrons. Many theories have been stated for explaining the structure of an atom.

Shape of the D Orbital

Shapes of orbitals are an approximate representation of boundaries in space for finding electrons occupied in that respective orbital. D orbitals are known to have a clover leaf shape or dumbbell inside where electrons can be found.

Question

Part C
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.10 x 1015HZ ?
Express your answer in joules to three significant figures.
• View Available Hint(s)
Ην ΑΣφ
?
X•10"
KE=
J
Submit
Previous Answers

I Rev
Electrons are emitted from the surface of a metal when it's exposed to
light. This is called the photoelectric effect. Each metal has a certain
threshold frequency of light, below which nothing happens. Right at
this threshold frequency, an electron is emitted. Above this frequency,
the electron is emitted and the extra energy is transferred to the
electron.
Here are some data collected on a sample of cesium exposed to various energies of light.
Light energy
(eV)
Electron KE
(eV)
Electron emitted?
3.87
no
The equation for this phenomenon is
3.88
no
KE= hv – hvo
3.89
yes
where KEis the kinetic energy of the emitted electron,
h = 6.63 x 10 34 J.s is Planck's constant, v is the frequency of the
3.90
yes
0.01
light, and vy is the threshold frequency of the metal.
3.91
yes
0.02
*Note that 1 eV (electron volt) =1.60 × 10 1º J.
Also, since E = hv, the equation can also be written as
KE=E-6
Part A
where E is the energy of the light and ø is the binding energy of the
electron in the metal.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Please answer question 12 Part A
Question: The energy of an electron transition in hydrogen atom is calculated using the following equation ΔΕ = -2.18 × 10-¹6/ n (a) Calculate the wavelength associated for the electron transition from n, = 3, 4, 5, 6, 7, 8 to n/= 2. (b) What part of the electromagnetic spectrum do these transitions belong? Hints: which transition corresponds to 410 nm? from n=5 to n=2 from n=3 to n=2 from n=6 to n=2 from n=7 to n=2 from n=8 to n=2 1. Plank's equation: E = 2 2. Use the following constants in separately as a reference cell: h= 6.626 x 10³4 J.s, c = 3 × 10 m/s 3. Construct a table like the following he n¡ n₁ AE λ(nm) ni 345678 222222 4 2 82
In the following equation what Ry? En =-R y 1 O the speed of light: 3.0 × 108 m/s O the number of photons O the principal quantum number a constant that is equal to 2.18 × 10-18 J O the energy of the electron in an orbit O the Rydberg constant: 1.097 × 107 m-¹
For photons with the following energies, calculate the wavelength and identify the region of the spectrum they are from: (a) 3.5 x 10−20 J, (b) 8.7 x 10−26 J, (c) 7.1 x 10−17 J, (d) 5.5 x 10−27 J
ch1
The photoelectric work function for a certain metal is 221.8 kJ/mol. If the metal absorbs a photon with a wavelength of 5.0 x 102 nm, and an electron is ejected, what is the kinetic energy of the ejected electron? (A) 4.0 x 10–19 J (B) 3.8 x 10–19 J (C) 7.7 x 10–20 J (D) 2.9 x 10–20 J
An atom of argon is moving in a container with a speed of 309.9 m/s. Determine its wavelength (in units of nm). Your answer must be in units of nanometers. molar mass atomic argon = 39.948 g/mol C = v) E = hv E=-(R/n?) X= (h/mu) AE = –RH [(1/n) – (1/n;²)] (1/A) = (Rµ/hc) [(1/n) – (1/n,?)] %3D %3D c = 2.9979 x 10³ m·s-1 h = 6.6260x 10 34 j.s %3D RH = 2.1798 x 1018 J 1 pm = 1x 10-12 m 1 nm = 1x 10-9m %3D 1 Å = 1x 10-10 m %3D NA = 6.0221 x 1023/ mol %3D
Find Zeff using Slater's Rules:
For Z effective, use Slater's Rules to find:
Which one is false?
(b) Photoelectron spectroscopy data for the 1s sublevel of Ne and the 1s sublevel of Xe are represented below. In terms of Coulomb's law and atomic structure, explain why the peak for Ne is positioned so far to the right of the peak for Xe. Ne 3500 3000 2500 2000 1500 1000 500 Binding Energy (MJ/mol) Хе 3500 3000 2500 2000 1500 1000 500 Binding Energy (MJ/mol) Relative Number Relative Number of Electrons of Electrons
The photoelectric work function for potassium metal is about 3.60 x 10–19 J. This is the minimum energy needed to eject an electron from the surface of potassium metal. What is the kinetic energy of an electron ejected by a photon of wavelength l (lamda) = 5.00 x 102 nm? (c = 3.00 x 108 m/s; h = 6.63 x 10-34 J.s) (A) 3.60 x 10–19 J (B) 3.98 x 10–19 J (C) 3.98 x 10–28 J (D) 3.78 x 10–20 J