Experiments have proved that sometimes light behaves like a wave, while in others, it behaves like particles. These particles of light are known as called photons. Photons can be thought of as both waves and particles. Louis de Broglie developed a formula to relate this dual nature as a wave and as well as a particle. It can also be applied to other particles such as electrons and protons. This formula relates the wavelength to the momentum of a wave or particle. In this article, we will discuss the De Broglie Wavelength formula with examples. Let us see it in detail.

**De Broglie Wavelength Formula**

**Introduction to the concept**

Louis-de-Broglie explained the concept of De-Broglie waves in the year 1923 which were later experimented and proved by Davisson and Germer in the year 1927. These waves explain the nature of the wave related to the particle.

For particles with mass like electrons, protons, etc., but not the photons, there is another form of the de Broglie wavelength formula. At non-relativistic speeds, the momentum of a particle will be equal to its rest mass m, multiplied by its velocity v.

The unit of the de Broglie wavelength is in meters. Since it is very small and hence expressed in nanometres or Angstroms units.

**Formula for De Broglie Wavelength:**

\(\lambda = \frac {h} {p}\)

Also,

\(\lambda = \frac {h} {m \; v}\)

Where,

\(\lambda\) | The de Broglie wavelength in meter |

h | Planck’s constant which is \(6.63 \times 10^ {-34} J s\) |

p | momentum of a particle in kg m \(s^{-1}\) |

m | mass of a particle in kg |

v | velocity of a particle m\( s^{-1}\) |

**Solved Examples**

Q.1: A certain photon has momentum \(1.50 \times 10 ^{-27 } kg m s^{-1}\). What will be the photon’s de Broglie wavelength?

Solution:

Known parameters given in the problem are:

\(p =1.50 \times 10 ^{-27 } kg m s^{-1}\)

And Plank’s constant, \(h = 6.63 \times 10^ {-34} J s\)

The de Broglie wavelength of the photon can be computed using the formula:

\(\lambda = \frac {h} {p}\)

= \(\frac { 6.63 \times 10^ {-34} } {1.50 \times 10 ^{-27 } }\)

= 4.42 \(\times 10^{-7} \)

= 442 \(\times 10 ^{-9}\)

= 442 Nano meter.

Therefore, the de Broglie wavelength of the photon will be 442 nm. This wavelength will be in the blue-violet part of the visible light spectrum.

Q. 2: The de Broglie wavelength of the electron is 0.26 nm. Electron particle has the mass of 9.109 \(\times 10^{-31}\) kg. Calculate the magnitude of the velocity of this electron.

Solution:

The known values given in the problem are:

m = 9.109 \(\times 10^{-31} kg \)

\(\lambda\) = 0.26 nm

The magnitude of the velocity of this electron can be found by using the formula with some rearrangement.

\(\lambda = \frac {h} {m \; v} \)

v = \(\frac {h} {m \lambda} \)

= \(\frac { 6.63 \times 10^{-34}} { 9.109\times10^{-31 }}\)

= 2.80 \(\times 10 ^6 m s^{-1} \)

Thus, the electron with de Broglie wavelength of 2.60 Armstrong will have the velocity of \(2.80 \times 10 ^6 m s^{-1}.\)

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…