In this article, we are going to ponder upon the energy and wavelength relationship along with examples and solve some problems to illustrate the same.

**Energy is directly dependent on the frequency of electromagnetic radiations. If the length of the wave increases, this implies that the recurrence of the wave will reduce that directly affecting the energy of the particle in a wave.**

**Energy and Wavelength Relationship Formula**

The energy of the particle can be related to its speed during propagation. The speed of the particle gives the idea of the frequency and the length of the wave. If the wavelength is minute then the frequency and hence the energy of the particle will increase.

**If the oscillations of the particle is more in a path trajectory, then the recurrence of the particle in a wave is more and wavelength is small, this implies that the energy possessed by the particle is more.**

The energy of any body is related to its wavelength by the equation

E=hc/λ

Where ‘h’ is a Planck’s constant h=6.626 *10^{-34}\ J.s

C is a velocity of light c=3 *10^{8} m/s and

λ is a wavelength of the light

**The energy is inversely proportional to the wavelength of the light. The lessened the wavelength, the more is the energy of the particle in a wave.**

**Problem 1: Calculate the energy of the photons emitting red light. Consider the wavelength of the beam of red light to be 698 nm. What will be the energy if the wavelength decreases to 500 nm that is if the source emits green light?**

**Given:**λ_{1}=698nm

λ_{2}=500 nm

h=6.626 *10^{-34} J.s

c=3 * 10^{8} m/s

We have,

E=hc/λ_{1}

E=6.626*10^{-34} J.s* 3 *10^{8} m/s/698* 10^{-9}m

=0.028* 10^{-17}=28* 10^{-20}Joules

The energy of the red wavelength is 28* 10^{-20}Joules.

If the wavelength λ_{2}=500 nm

Then the energy associated with the green light is

E=hc/λ_{2}

E=6.626*10^{-34} J.s* 3 *10^{8} m/s / 500* 10^{-9}m

=0.03910^{-17}=39* 10^{-20}Joules

**We can see that, the energy has increased to 39*10 ^{-20 }Joules as the wavelength is reduced.**

**Read more on Effect Of Refraction On Wavelength: How, Why, Detailed Facts.**

**Energy and Wavelength Relationship Graph**

As the wavelength increases, the frequency of the wave falls off thus declining the energy possessed by the wave. If we plot a graph of Energy v/s Wavelength of the emerging particle, then the graph will look like as shown below

**The above graph clearly indicated that, as the wavelength increases, the energy associated with the particle decreases exponentially.**

**Kinetic Energy and Wavelength Relationship**

**If the speed of the particle is greater, it is evident that the kinetic energy of the particle is high.** The kinetic energy is given by the equation

K.E=1/2mv^{2}

Where m is a mass of the object or particle

V is a velocity of the mass

We can write the above equation as

2E=mv^{2}

Multiplying ‘m’ on both sides of the equation

2mE=(mv)^{2}

**The momentum of the object is given as the product of object mass and the velocity at which it is moving.**

p=mv

Hence, the above equation becomes

P^{2}=2mE

P=√2mE

According to De Broglie,

λ =h/p

Substituting the above equation, we have

λ =h/ √2mE

The above equation gives the relation between the energy and the wavelength of the particle.

**Read more on What Is The Kinetic Energy Of Light:Detailed Facts.**

**Problem 2: Calculate the kinetic energy of a particle of mass 9.1 × 10**^{-31 }kg having a wavelength 293nm. Also, find the velocity of the particle.

^{-31 }kg having a wavelength 293nm. Also, find the velocity of the particle.

**Given:** λ =293 nm

m = 9.1 × 10^{-31 }kg

h=6.626 *10^{-34}J.s

c=3 *10^{8} m/s

We have,

λ =h/ √2mE

λ^{2}=h^{2}/ 2mE

E= h^{2}/ 2mλ^{2}

=(6.626 * 10^{-34} J.s)^{2}/2* 9.1* 10^{-31}* (293*10^{-9}) ^{2}

=0.28*10^{-23}

The kinetic energy associated with the particle is 0.28* 10^{-23} Joules.

Now, to calculate the velocity of the particle, let us derive a formula for velocity from the kinetic energy,

K.E=1/2 mv^{2}

2E= mv^{2}

v=√(2E/m)

= √(2(0.28*10^{-23})/(9.8*10^{-31}))

=0.24*10^{4}=2400m/s

The velocity of the particle having wavelength 298 nm is 2400 m/s.

**Electron Energy and Wavelength Relationship**

The energy of the electron is given by the simple equation as

E=h\nu

Where ‘h’ is a Planck’s constant and

nu is a frequency of occurrence of the electron

The frequency of the electron is given as

nu =v/λ

Where v is a velocity of the electron and

λ is a wavelength of the electron wave

Hence, the energy is related to the wavelength of the electron as

E=hv/λ

**This is a relation to find the energy associated with the single electron propagating having a specific wavelength, speed, and frequency. The energy is inversely proportional to the wavelength. If the wavelength of the electron is lessened, the energy of the wave has to be greater.**

Upon receiving the energy in some form, the electron gets excited from the lower energy state to the higher energy state. For the transition of the electrons from one state to another, the energy of the electron is given by the equation

E=R_{E}(1/n_{f}– 1/n_{i})

Where R_{E}=-2.18* 10^{-18}m^{-1} is a Rydberg Constant

n_{f} is a final state of the electron

n_{i} is the initial state of the electron

We can further rewrite the above equation as

h\nu =R_{E}(1/n_{f}– 1/n_{i})

hc/λ =R_{E}(1/n_{f}– 1/n_{i})

1/λ =R_{E}hc(1/n_{f}– 1/n_{i})

1/λ =R(1/n_{f}– 1/n_{i})

Where,

R=R_{E}hc=1.097* 10^{7}

As the electron gains the energy, the electron transit and jumps in the higher state of energy level and releases the energy to the electrons present in that state and either gets stable or releases the amount of energy and returns down to the lower energy states.

**Read more on 16+ Amplitude of a wave example: Detailed Explanations.**

**Problem 3: If the electron transit from state n**_{i}=1 to state n_{f}=2, then calculate the wavelength of the electron.

_{i}=1 to state n

_{f}=2, then calculate the wavelength of the electron.

**Given:**

n_{i}=1

n_{f}=2

1/λ =R_{E}(1/n_{f}– 1/n_{i})

1/λ=-1.097*10^{7} * ( 1/2-1/1 )

1/λ=0.5485* 10^{7}

Hence,

λ =1/0.5485* 10^{7}

λ =1.823*10^{-7}

λ =182.3*10^{-9}=182.3nm

The wavelength of the light emitted during the transition of electron from one energy level to the other is 182.3 nm.

**Radiant Energy and Wavelength Relationship**

Every object absorbs light rays during daylight depending upon its shape, size, and its composition. If the temperature of the surface of the object reaches above the absolute zero temperature, the object will emit the radiations in the form of waves.

**This emitted radiation is proportional to the fourth power of the absolute temperature of the object and is given by the equation**

U=ɛΣ T^{4A}

Where U is a radiated energy

ɛ is the emissivity of radiation from the object

Σ is the Stefan-Boltzmann constant and is equal to Σ=5.67*10^{-8}W/m^{2}K^{4}

T is an absolute temperature

A is the area of the object

**The object at high temperature emits radiation of short wavelengths, and the cooler surfaces emit waves of large wavelength.** Based on the emission of radiation, and the wavelength of the emitted radiations, the waves are classified as given in the below chart.

Name |
Radio Waves |
Microwaves |
Infrared |
Visible |
Ultra Violet |
X-rays |
Gamma rays |

Wavelength |
>1m | 1mm-1m | 700nm-1mm | 400nm-700nm | 10nm-380nm | 0.01nm-10nm | <0.01nm |

Frequency |
<300MHz | 300MHz-300GHz | 300GHz-430THz | 430THz-750THz | 750THz-30PHz | 30PHz-30EHz | >30Ehz |

**As the wavelength of the radiations decreases, the frequency of the wave escalates. Wavelength is directly related to the temperature, hence if the frequency of the emitted radiation is more this implies that the energy of the object is high.**

The gamma rays, x-rays, and UV rays have a very short wavelength, hence the energy of these waves is very high compared to visible, infrared, microwaves, or radio waves. Also, the higher is the radiations received by the object, the more it will emit out depending upon the emissivity factor of the object.

Below is a graph of energy v/s wavelength plotted for different temperatures. The graph shows that, as the temperature of the system rises, the energy of the emitted radiations also increases with temperature.

For, the wavelength in the visible region, the emission of radiation is the maximum. This is because the Sun emits UV rays along with infrared rays and visible rays and the rays are the electromagnetic waves for long range. The ozone layer of the Earth shields the earth’s atmosphere from this harmful radiation and gets either reflected back or trapped in the clouds.

**More of the radiations are emitted in the visible region during day time as more and more radiations are received during day time from the Sun and fewer IR rays are given out compared to the visible spectrum. At night, the temperature goes down, the wavelength of the radiation increases, and more of the IR rays are emitted by the object.**

**Read more on Properties Of Refraction: Wave, Physical Properties, Exhaustive Facts.**

**Problem 4: A box length 11cm, width 2cm, and breath 7cm is heated at 1200 Kelvin temperature. If the emissivity of the box is 0.5, then calculate the rate of radiation of energy from the box.**

**Given:**l=11cm

h=2cm

b=7cm

ɛ =0.5

Σ=5.67* 10^{-8}W/m^{2}K^{4}

T=1200 K

The total surface area of the box is

A=2(lb+bh+hl)

=2(11*7+7*s 2+2*11)

=2 (77+14+22)

=0.0226 sq.m

The energy radiated by the box is

U=ɛ Σ T^{4A}

=0.5* 5.67* 10^{-8}* 1200^{4}* 0.0226

=1328.6 Watts

**Energy Frequency and Wavelength Relationship**

**If the frequency of the wave is more, the energy associated with the particle is more. The energy is related to the frequency of the wave as**

E=h/nu

Where ‘h’ is a Planck’s Constant

nu is a frequency of the wave

**The frequency of the wave is defined as the speed of the wave in the medium and the wavelength of the wave.**

nu =v/λ

Where v is a velocity of the wave

λ is a wavelength

Hence,

λ=v/nu

This gives the relation between the frequency and wavelength of the wave. It says that as the wavelength and frequency both are inversely correlated with each other. If the wavelength increases, the frequency f the wave will decrease.

**Read more on Effect Of Refraction On Frequency: How, Why Not, Detailed Facts.**

**Problem 5: The speed of a beam of light emitted from the source is 1.9 × 10**^{8 }m/s. The frequency of occurrence of the emitted wave is 450THz. Find the wavelength of the emitted radiation.

^{8 }m/s. The frequency of occurrence of the emitted wave is 450THz. Find the wavelength of the emitted radiation.

**Given:** v=1.9*10^{8} m/s

F=450THz=450*10^{12}Hz

The wavelength of the beam of light is

λ =v/f

=1.9* 10^{8}/ 450* 10^{12}

=0.004222*10^{-4}

=422.2* 10^{-9}=422.2nm

The beam of light is of wavelength 422.2 nm.

**Energy of Photon and Wavelength Relationship**

**The energy possessed by a photon is termed as photon energy, and is inversely proportional to the electromagnetic wave of the photon, by the relation**

E=hc/λ

Where ‘h’ is a Planck’s constant

C is a speed of light

λ is a wavelength of the photon

The frequency of the photon is given by the equation

f=c/λ

Where f is a frequency

Hence, the photon with a large wavelength possesses a small unit of energy whereas the photon with a smaller wavelength gives a large amount of energy.

**Read more on What is the wavelength of photon: How to Find, Several Insights And Facts.**

**Problem 6: Calculate the energy of the photon propagating in an electromagnetic wave having a wavelength of 620 nm.**

**Given:** Wavelengthλ =620 nm

h=6.626 *10^{-34} js

c=3 *10^{8} m/s

We have,

E=hc/λ

E=6.626 * 10^{-34} J.s*3 * 10^{8 }m/s/620* 10^{-9}m

=0.032*10^{-17}=32*10^{-20 }Joules

The energy associated with the photon is 32* 10^{-20}Joules.

**Frequently Asked Questions**

**Q1. Calculate the wavelength of the electron traveling at a speed of 6.35 × 10**^{6 }m/s

^{6 }m/s

**Given:** v=6.35*10^{6}m/s

m=9.1*10^{-31}kg

h=6.62* 10^{-34 }J.s

The kinetic energy of the electron is

K.E=1/2 mv^{2}

=1/2 * 9.1*10^{-31}* (6.35* 10^{6})^{2}

=1.83* 10^{-17}Joules

The momentum of the electron is

P=√2mE

=√2* 9.1* 10^{-31}*1.83* 10^{-17}

=5.7*10^{-24}kg.m/s

Now, the wavelength of the electron is

λ =h/√2mE

=6.62*10^{-34}/5.7*10^{-24}

=4.8*10^{-10}m

=48nm

The wavelength of the electron moving with velocity 6.35*10^{6}m/s is 48 nm.

**Q2. A black object having a surface area of 180 sq.m is kept at a temperature of 550K. What is the rate of radiation of energy from the object?**

**Given:** A=180 sq.m

T=550K

Since the object is black in color, the emissivity is 1.

ɛ =1

We have,

U=ɛΣT^{4A}

=1*s 5.67* 10^{-8}*550^{4}*180

=0.93*10^{6}Watts

The power radiated from the emission of radiation from the object is 0.93*10^{6}Watts.

**What is the absolute temperature of the system?**

It is a non-variable and a perfect value of the temperature of a system.

**The absolute temperature of the system is measured on the scale of degree Celsius, Fahrenheit, or Kelvin which measure zero as absolute zero degrees.**

**How does the wavelength of the photon be dependent on the temperature?**

The temperature of the system specifies the agility of the particles of the system.

**The more radiations received by the system at higher temperatures, the more emission will be given out from the system. At higher temperatures, shorter wavelength radiations are omitted and at lower temperatures, longer wavelengths are radiated.**

**Also Read:**

- Conservation of mechanical energy examples
- Electrical energy to chemical energy
- Example of radiant energy to kinetic energy
- Example of gravitational energy to mechanical energy
- Examples of potential energy in your home
- Mechanical energy to chemical energy
- Example of radiant energy to mechanical energy
- What does not affect potential energy
- Can potential energy be negative
- Does velocity affect potential energy

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess.