It is a PowerPoint presentation on numerical problems of Planck quantum theory topic. It includes 12 numerical problems with solved answers. Besides, this PowerPoint presentation has a mind map to remember all formulas of Planck quantum law.

1. By,
Jayam chemistry adda

2. Planck quantum formula
(𝐄 = 𝐡𝛎)
For a single photon For “n” no of photons
(𝐄 = 𝐧𝐡𝛎)
𝐄 =
𝐡𝐜
𝛌
Planck wavelength equation
𝐄 = 𝐡𝐜Ῡ
Planck wavenumber equation
(Or)
𝑬 =
𝟏𝟗. 𝟖𝟕𝟖 × 𝟏𝟎−𝟐𝟔
𝝀
(Or)
𝐄 = (𝟏𝟗. 𝟖𝟕𝟖 × 𝟏𝟎−𝟐𝟔
) × Ῡ
𝐸 =
(6.626 × 10−34
) × (3 × 108
)
λ
𝐸 = (6.626 × 10−34) × (3 × 108) × Ῡ
Units of Planck quantum
SI Unit---- Joule
CGS unit--- Erg
Atomic unit ---
electron volt (eV)
Each discrete tiny packet
of energy is quantum.
And the bundle of light
energy is called photon.
In 1900, Max Planck explained
the particle nature of light with
his theory known as Planck's
quantum theory.
Jayam chemistry adda
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Problem-1: Calculate the energy of a photon at 525 nm.
The Wavelength of photon=525 nm
Solution:
The formula to measure the energy of photon is;
𝐄 =
𝟏𝟗. 𝟖𝟕𝟖 × 𝟏𝟎−𝟐𝟔
𝛌
E=
19.878×10−26
525×10−9 joule
E=0.0378×10−17 joules
𝑬 = 𝟑. 𝟕𝟖 × 𝟏𝟎−𝟏𝟗
𝒋𝒐𝒖𝒍𝒆𝒔

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Problem-2: The energy absorbed by each molecule (A2) of a compound is 8.4 X 10-19 joules.
And bond energy per molecule is 8.0 X 10-19 joules. What is the kinetic energy per atom of
the molecule?
Energy absorbed by a molecule = 8.4 X 10-19 joules
Solution:
Bond energy of each molecule of the substance = 8.0 X 10-19 joules
The formula to calculate the kinetic energy per atom of the molecule is;
𝐾𝐸 =
8.4 × 10−19
− 8.0 × 10−19
𝑗𝑜𝑢𝑙𝑒𝑠
2
Kinetic energy of each atom of a molecule =
Energy absorbed − (bond energy per molecule)
number of atoms in the molecule
KE = 0.2 × 10−19 joules = 2 × 10−20 joules

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Problem-3: Find the number of photons of light at 4000 pm having 1 joule of energy ?
Solution:
Energy of light = 1 joule
Wavelength of light = 4000 pm= 4000 x 10-12 m
The formula to measure the number of photons of light is;
n =
λ𝐸
ℎ𝑐
=
λ𝐸
19.878 × 10−26 𝑗𝑚
E = nhν =
nhc
λ
(By substituting the values of h and c)
𝑛 =
4000 × 10−12
𝑚 × 1 𝑗
19.878 × 10−26 𝑗𝑚
𝑛 = 201.22 × 1014 = 2.0122 × 1016

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Problem-4: A 242 nm electromagnetic light can ionize a sodium atom. What is the ionization
energy of sodium in KJ/mole?
Solution:
The wavelength of electromagnetic light = 242 nm = 242 x 10-9 m
The formula to calculate the ionization energy of one sodium atom is;
E =
19.878 × 10−26
Jm
242 × 10−9 m
E = 0.0821 × 10−17
Joule
𝐸 =
19.878 × 10−26
Jm
λ
Ionization energy of one mole of a substance = ionization energy of one atom x Avogadro's number
Ionization energy of one mole of sodium = (0.0821 × 10−17
joule) × (6.023 × 1023
)
Ionization energy of one mole of sodium = 0.494 × 106
joule/mole = 494 KJ/mole

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Problem-5: A nitrogen laser produces electromagnetic radiation at 337.1 nm and emits 5.6 X 1024
photons. Calculate the energy of the radiation.
Solution:
Wavelength of electromagnetic radiation = 337.1 nm = 337.1 x 10-9 m
Number of photons emitted by the laser = 5.6 X 1024
The formula to calculate the energy of radiation is;
E =
nhc
λ
=
n 19.878 × 10−26 jm
λ
E =
5.6 × 1024
× 19.878 × 10−26
jm
337.1 × 10−9 m
E = 0.3302 x 107 joules
E = 3.302 × 106
joules

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Problem-6: Calculate the energy of the photon having frequency of 4.5 X 1012 Hz.
Solution:
The frequency of photon = 4.5 X 1012 Hz
The formula to measure the energy of photon is:
E = hν
E = 6.626 × 10−34 js × 4.5 × 1012 Hz
𝐸 = 29.8 × 10−22 joule
E = 2.98 × 10−21 joule

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Problem-7: Find the distance traveled by light in a vacuum in 20 seconds.
Solution:
Time period of light in vacuum= 20 seconds
Velocity of light in vacuum = 3 x 108 m/sec
The formula to calculate the distance travelled by light in vacuum= velocity x distance
Distance travelled by light in vacuum in 20 seconds is;
d= 6 x 109 m
𝑑 = 3 × 108
𝑚
𝑠𝑒𝑐
× 20 𝑠𝑒𝑐

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Problem-8 : Calculate the energy of a photon whose frequency is 6 x 1020 Hz
Solution:
The frequency of photon = 6 x 1020 Hz
The formula to calculate the energy of photon is;
E = hν
E = 6.626 × 10−34
js × 6 × 1020
Hz
E = 39.75 × 10−14
joules
1 𝑒𝑉 = 1.602 × 10−19 𝑗𝑜𝑢𝑙𝑒
E =
39.75 × 10−14
1.602 × 10−19 eV
E = 24.8 × 105 eV = 2.48 × 106 eV = 2.48 MeV

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Problem-9: What is the energy of a mole of photons having a wavelength of 715 nm?
Solution:
The wavelength of photon=715 nm=715 x 10-9 nm
The formula to measure the energy of one mole of photons is;
𝐸 =
19.878 × 10−26 𝑗𝑜𝑢𝑙𝑒𝑠
715 × 10−9 × (6.023 × 1023
)
𝐸 =
19.878 × 10−26
𝑗𝑜𝑢𝑙𝑒𝑠
λ
× 𝐴𝑣𝑜𝑔𝑎𝑟𝑑𝑜′
𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
𝐸 = 0.0278 × 10−17
𝑗𝑜𝑢𝑙𝑒𝑠 × (6.023 × 1023
)
E=0.1674 x 106 joules
E = 1.674 x 105 joules

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Problem-10: Electromagnetic radiation has a frequency of 8.3 x 1014 Hz. Then find its energy in
eV.
Solution:
The frequency of electromagnetic radiation = 8.3 x 1014 Hz
The formula to calculate the energy of electromagnetic radiation is;
E = hν
h = 4.136 x 10-15 eV second
E = 4.136 × 10−15 eV second × 8.3 × 1014 Hz
𝐸 = 34.32 × 10−1
eV
E = 3.432 eV

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Problem-11: The value of Planck’s constant is 6.626 x 10-34 Js. The velocity of light in a
vacuum is 3 x 108 m/sec. What is the wavelength of light at 8 x 1015 Hz?
Solution:
The frequency of light = 8 x 1015 Hz
The formula to calculate the wavelength of light is;
λ =
c
ν
λ =
3 × 108 m/sec
8 × 1015 Hz
λ = 0.375 × 10−7 m
λ = 37.5 × 10−9
m = 37.5 nm
Where, (c is velocity of light in vacuum)

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Problem-12: The energies of the two photons are E1 and E2. The wavelengths associated with the
two photons are 9000 nm and 18000nm. What is the relationship between E1 and E2?
The wavelength of photon-1=λ1= 9000nm
The wavelength of photon-2=λ2= 18000 nm
By Planck quantum law, the relationship between photon’s energy and its wavelengths is
E =
hc
λ
E ∝
1
λ
𝐸1
𝐸2
=
λ2
λ1
E1
E2
=
18000
9000
= 2
E1 = 2E2
Solution: