Planck constant-MCQs & answers-Jayam chemistry learners
MCQs with answers on Planck constant
Introduction:
 |
| MCQs of Planck's constant |
Table of contents:
Multiple choice questions and answers of Planck’s constant
Define Planck's constant
Mind map of Planck's constant:
Multiple choice questions and answers of Planck's constant:
1. Planck quantum theory laid foundation to___________
(i) Classical physics
(ii) Quantum mechanics
(iii) Thermodynamics
(iv) Spectroscopy
Answer: Quantum mechanics
Explanation:
It explains the particle nature of light on the atomic and sub-atomic levels. Thus, Planck's constant plays a vital role in quantum mechanics.
2. What is the unit for angular frequency?
(i) Radians per second
(ii) Hertz
(iii) Second
(iv) Meter
Answer: Radians per second
Explanation:
Angular frequency measures angular displacement in unit time. The unit of angular frequency is radians per second
(i) 16.7
(ii) 6.626
(iii) 19.8 x 10-26
(iv) 5
Answer: 19.8 x 10-26
Explanation:
The relationship between photon's energy and wavenumber by Planck quantum theory is;
E = hcῩ
E = (19.878 x 10-26 joule
meter) x wave number
(i) P
(ii) h
(iii) k
(iv) B
Answer: h
Explanation:
The symbol "h" denotes Planck constant
5. What is Bohr’s
quantized angular momentum condition?
(i) mvr=nħ
(ii) mvr=nh
(iii) mvr=nπ
(iv) mvr=hπ
Answer: mvr=nħ
Explanation:
Neil Bohr introduced the reduced
Planck's constant to express quantized angular momentum of electron orbits.
mvr = nħ
(i) 900 Hz
(ii) 25 Hz
(iii) 2.5 X 107
Hz
(iv) 250
Answer: 2.5 X 107
Hz
Explanation:
According to Planck’s quantum theory, the relationship between wavelength and frequency of light is ν = c/λ
ν = 3 x 108 /12 =0.25 x 108
Hz
ν = 2.5 x 107 Hz
7. Which of the following physical quantity has the same units as that of reduced Planck's constant?
(i) Angular
frequency
(ii)
Frequency
(iii)
Angular momentum
(iv)
Angular velocity
(i) de-Broglie wavelength
(ii) Bohr electron energy
(iii) Photoelectric effect
(iv) Magnetic moment of the electron
Answer: Magnetic moment of orbital
Explanation:
The formula for magnetic moment of electron is μ = [n (n+2)]½
The
magnetic moment formula of the electron involves the principal quantum number.
And it is independent of the value of Planck's constant.
(i) 1012 Hz
(ii) 105 Hz
(iii) 1011 Hz
(iv) 108 Hz
Answer: 1011 Hz
Explanation:
According to Planck’s quantum theory, the relationship between wavelength and frequency of light is ν = c/λ
ν = 3 x 108 /3 x 10-3
10. Einstein-Planck's law describes_______________
(i) The size of a photon
(ii) The number of photons
(iii) The wavenumber of photon
(iv) The angular frequency of photon
Answer: The size of a photon
Explanation:
The Einstein-Planck's law is
E=hf
Where,
E= Energy of photon
h= Planck's constant
f= frequency of photon
It determines the energy of the photon. In other terms, the size of the photon.
11. What is the energy of light having a 9 nm wavelength?
(i) 2.2 X
10-17 J
(ii) 4.5 X
1923 J
(iii) 6.4 X
10-26 J
(iv) 3.8 X
10-17 J
Answer: 2.2
X 10-17 J
Explanation:
The
wavelength of light= 9 nm= 9X 10-9 m
The formula
to calculate the energy of photon is
E = 19.878 x 10-26 /λ Joules
E = 19.878 x 10-26 /9 x 10-9
J
E = 2.208 x 10-17 J
12. Which of the following experimental method used to determine Planck’s constant in laboratories?
(i) Particle accelerator method
(ii) Ion-exchange method
(iii) Vapor density method
(iv) Determination of the mass of the sample
Answer: Particle accelerator method
Explanation:
The following are a few experimental methods to compute
Planck's constant practically.
1. From Faraday's constant in electrolysis experiments
2. Particle accelerator method
3. Kibble balance
13. What is the ratio of the energy of photons having frequencies 4000Hz and 8000 Hz?
(i) 2
(ii) ½
(iii) 3.14
(iv) 4
Answer: 1/2
Explanation:
According
to Planck’s quantum theory,
E1/E2 = ν1/ν2
E1/E2 = 4000 Hz/8000
Hz
E1/E2 = 1/2
14. Why are LED lights used to calculate Planck constant?
(i) It is a reliable light source available at a cheaper
cost
(ii) It emits light at specific wavelengths
(iii) It emits different colored radiations at different
threshold voltages
(iv) It transfers electric energy to light radiation
Answer: It emits different colored radiations at different
threshold voltages
Explanation:
LED (light emitting
diodes) helps to enumerate Planck constant experimentally.
It is due to its ability to emit different colored
radiations at different threshold voltages while producing electrons.
15. The speed of the electromagnetic wave in a vacuum is_____________
(i) 6.626 x 10-27
(ii) 3 x 108
(iii) 6.023 x 1023
(iv) 5 x 1027
Explanation:
The electromagnetic wave travels with a constant speed of 3 x 108 m/sec in a vacuum.
16. Planck’s constant is _________________
(i) Proportionality constant
(ii) Universal constant
(iii) Fundamental constant
(iv) All of the above
Answer: All of the above
Explanation:
It shows the proportionality relation of Planck's constant
with the photon's energy and frequency. Hence, it is a proportionality
constant.
It is a fundamental physical quantity. Its magnitude and
unit are independent of other physical quantities for measurements.
It is a constant value that does not change with time. So,
it is a universal constant.
17. Why did Planck’s constant introduce in the quantum equation?
(i) Chemical equations must have a constant value
(ii) Planck felt to introduce it to have his name
(iii) It describes the photon energy from the frequency of
light
(iv) A constant only can express the energy of light
Answer: It describes the photon energy from the frequency of
light
Explanation:
The classical physics depictions of light could not explain
phenomena such as black body radiations and the photoelectric effect.
But, the limitations of classical physics in explaining the
ultraviolet catastrophe of black bodies laid the foundation for Planck’s
experiments on the energies of oscillating particles.
Planck’s quantum theory proposed the quantum nature of
electromagnetic radiant energies.
When Planck researched black body emissions in 1900, he
found that a proportionality constant is essential in his empirical formula to
match the experimental results.
With effort, he calculated the value of h to measure the
quantum energy.
Max Planck received the Nobel Prize in physics in 1918 for
his work on the energy quanta.
18. The Planck’s constant applies to______________
(i) Microscopic particles
(iii) Macroscopic objects
(iii) Both options (a) and (b)
(iv) Neither (a) nor (b)
Answer: Microscopic particles
Explanation:
Planck’s constant explains the particle nature of light on
the atomic and sub-atomic levels. Thus, it plays a vital role in quantum
theory.
19. Unit of
Planck’s constant in atomic units is _________________
(i)
eV/hertz
(ii) eV
hertz
(iii) Joule
second
(iv)
Joule/hertz
Answer: eV/hertz
Explanation:
In case we express
energy in atomic units, the unit of Planck's constant is eV hertz-1
or eV second.
20. What is the influence of time on Planck's constant?
(i) It varies directly with time
(ii) It varies inversely with time
(iii) It depends on geographical location
(iv) It remains constant
Answer: It remains constant
Explanation:
h is a constant value that does not change with time. So, it is a universal constant.
Match the following quiz:
| Column-A | Column-B |
|---|---|
| A. Rydberg constant | 1. Maximum intense wavelength of blackbody emissions |
| B. Planck's constant | 2. Limiting value of highest wavenumber of photon |
| C. Wien constant | 3. Size of energy chunks |
| D. Stefan's constant | 4. Quantized angular momentum of electron orbits |
| E. Reduced Planck's constant | 5. Amount of heat radiated from blackbody |
Answers to the above match the following table is A-2, B-3, C-1, D-5, and E-4.
Numerical problems on Planck's constant:
Question-1: What is
the threshold energy of light radiation whose wavelength is 200 nm? If the
kinetic energy of the ejected electron is 1.68 x 106 J/mol.
Answer:
Wavelength of emitted
electromagnetic radiation = 200 nm = 200 x 10-9 m
According to Planck
quantum law, we have;
E = hc/λ
h = Planck’s constant
= 6.626 x 10-34 joule seconds
c = Velocity of light
in a vacuum = 3 x 108 m/sec
hc = (6.626 x 10-34
joule seconds) x (3 x 108 m/sec)
hc = (19.878 x 10-26
joule meter)
E = (19.878 x 10-26
Jm) / 200 x 10-9 m
E = 0.09939 x 10-17
joule
Energy of one mole of
photon = (Energy of single photon) x Avogadro’s number
Energy of one mole of
photon= (0.09939 x 10-17) x (6.023 x 1023)
Energy of one mole of
photon = 0.5986 x 106 joule/mole
By photoelectric
effect formula, we have;
½mv2 = hν-hν0
According to question,
the kinetic energy of the ejected electron = 1.68 x 106 joule/mole
By substituting the
values in the above equation, we get;
1.68 x 106
joule/mole = 0.5986 x 106 joule/mole – hν0
hv0 = -1.0814
x 106 joule/mole
Question-2: What is
the frequency of electromagnetic radiation if the energy difference is 30 x 10-34
joule?
Answer:
The energy difference
of photon = 30 x 10-34 joule
Following the Planck’s
quantum formula, we have;
ΔE = hν
By substituting the
values of ΔE and h, we get;
(30 x 10-34
joule) = (6.626 x 10-34 joule seconds) x ν
ΔE = 4.527 sec-1
Question-3: Calculate
the emission rate per second of released electromagnetic radiation whose
wavelength is 0.5 nm. And the power of the bulb emitting light is 20 watts.
Answer:
Wavelength of emitted
electromagnetic radiation = 0.5 nm = 0.5 x 10-9 m
According to Planck
quantum law, we have;
E = hc/λ
h = Planck’s constant
= 6.626 x 10-34 joule seconds
c = Velocity of light
in a vacuum = 3 x 108 m/sec
hc = (6.626 x 10-34
joule seconds) x (3 x 108 m/sec)
hc = (19.878 x 10-26
joule meter)
E = (19.878 x 10-26
Jm) / 0.5 x 10-9 m
E = 39.756 x 10-17
joules
Rate of emission of
photon per second = power of the bulb / energy of the emitted photon
According to the
question, the power of electric bulb = 20 watts = 20 joule/seconds
By substituting the
values in the above equation, we get;
Rate of emission of
photon per second = 20 / 39.756 x 10-17 sec-1
Rate of emission of
photon per second = 0.5030 x 1017 sec-1
Question-4: If the
work function of a metal surface is 2 eV. What is the threshold wavelength of
the light radiation?
Answer:
The work function of a
metal surface = 2 eV
hν = 2 eV = 2 x (1.602
x 10-19 joules)
hν = 3.204 x 10-19
joules
According to the
relationship between the wavelength and frequency, we have;
ν = c/λ
hν = hc/λ = 3.204 x 10-19
joules
h = Planck’s constant
= 6.626 x 10-34 joule seconds
c = Velocity of light
in a vacuum = 3 x 108 m/sec
hc = (6.626 x 10-34
joule seconds) x (3 x 108 m/sec)
hc = (19.878 x 10-26
joule meter)
By substituting the
values in the above equation, we get;
(19.878 x 10-26
joule meter) /λ = 3.204 x 10-19 joules
λ = (19.878 x 10-26 joule meter)
/ 3.204 x 10-19 joules
λ = 620 nm
Question-5: What is
the threshold frequency of the metal if the electron’s binding energy is 193
J/mole?
Binding energy of one
mole of photon =hv0 = 193 J/mole
Binding energy of single electron = 193 / (6.023
x 1023) J = 32.04 x 10-23 J
E= hv0
ν0 = E / h = 32.04 x 10-23 J / 6.626 x 10-34
Js
ν0 = 4.835 x 1011 s-1
Question-6: What is the
mass of photon of sodium having wavelength of 550 nm?
Answer:
Wavelength of sodium
photon = 550 nm
By following
de-Broglie equation, we have;
λ = h/mv
m = (6.626 x 10-34
js) / (3 x 108 m/sec) (550 x 10-9 m)
m = 0.00401 x 10-33
Kg
m = 4.01 x 10-36
Kg
Question-7: What is
the wavelength of the ball having 0.1 kg mass moving with a velocity of
20m/sec?
Mass of ball = 0.1 kg
Velocity of ball = 20
m/sec
According to
de-Broglie law, the formula for calculating the wavelength of photon is λ = h/mv
λ = (6.626 x 10-34 joule second)
/ 0.1 Kg x 20 m/sec
λ = 3.313 x 10-34 seconds
Question-8: A gas
absorbs a photon of 100 nm and emits two spectral lines. The wavelength of one
emission line is 125 nm. What is the wavelength of the other spectral line?
Answer:
E1 is the energy
of the photon absorbed. E2 and E3 are the spectral lines energy
emitted.
Then E1 = E2+E3
According to Planck
quantum theory,
ΔE = hc/λ
hc/λ1 = hc/λ2 + hc/λ3
1/100 = 1/125 + 1/λ3
1/λ3 = 1/100
-1/125
λ3 = 500 nm
Question-9: What is
the energy of the photon corresponding to the light frequency 5 x 1015
sec-1?
Answer:
The frequency of the
electromagnetic radiation = 5 x 1015 sec-1
According to Planck
quantum theory, we have;
E = hν
E = (6.626 x 10-34
joule second) x (5 x 1015 sec-1)
E = 33.13 x 10-19
joules
Question-10: What is
the angular momentum of the hydrogen electron in the fourth orbit?
Answer:
According to Bohr
angular momentum condition mvr = nh/2π
mvr = 4h/2π
mvr = 2 x (6.626 x 10-34
joule second) / 3.14
mvr = 4.22 x 10-34 joule second
Planck's constant and reduced Planck's constant:
Students always confuse with h and ħ, which are the symbols that designate Planck's constant and reduced Planck's constant.
Max Planck discovered Planck's constant to enumerate the size of quantum. Besides, Neil Bohr invented reduced Planck's constant to calculate the angular momentum of quantized Bohr orbits.
Both h and ħ are numerical physical quantities having values 6.626 x 10-34 joule seconds and 1.054 x 10-34 joule seconds.
Planck's constant defines linear frequency. But, reduced Planck's constant shows angular frequency.
Planck's constant is the ratio of photon magnitude to the radiation frequency. And the reduced Planck's constant is the ratio of Planck's constant and 2π.
The natural unit of h is 2π, and that of ħ is 1.
Our effort to present
the best content on multiple choice questions and answers to Planck quantum
theory topic might be helpful to you. We suggest you check the external sources
linked with the blog post for additional knowledge.