Solved Examples on De-Broglie Wavelength
Example 1: What is the wavelength of an electron moving at 5.31 x 106 m/sec?
Solution:
Given:
mass of electron = 9.11 x 10-34 kg
h = 6.626 x 10-34 J·sde Broglie’s equation is
λ = h/mv
λ = 6.626 x 10-34 J·s/ 9.11 x 10-31 kg x 5.31 x 106 m/sec
λ = 6.626 x 10-34 J·s / 4.84 x 10-24 kg·m/sec
λ = 1.37 x 10-10 m
λ = 1.37 Å
The wavelength of an electron moving 5.31 x 106 m/sec is 1.37 x 10-10 m or 1.37 Å.
Example 2: What is the de Broglie wavelength of a 0.05 eV neutron?
Solution:
λ = h / p
= h / √2mok
= hc / √(2moc2 )K
= 12.4 x 103 / √2(940 x 106) (0.05)λ = 1.28 Å
The de Broglie wavelength of a 0.05 eV(thermal) neutron is 1.28Å.
Example 3: A certain photon has momentum 1.50×10-27kgms-1. What will be the photon’s de Broglie wavelength?
Solution:
p = 1.50×10-27kgms-1
h = 6.63×10-34Js
De Broglie wavelength of the photon can be computed using the formula:
λ = h / p
= 6.63×10 -34 / 1.50×10-27
= 4.42 ×10-7
= 442 ×10-9
= 442 Nanometer.
Therefore, the de Broglie wavelength of the photon will be 442 nm
Photocell and Probability Interpretation to Matter Waves
Dual nature of radiation and matter is a fundamental concept in physics that describes the wave-like and particle-like behaviour of both radiation and matter. This concept was first introduced by Louis de Broglie in 1924, who proposed that all matter, not just electromagnetic radiation, has wave-like properties. This idea was later confirmed by the famous double-slit experiment and the development of quantum mechanics.
Let’s learn about DeBroglie’s Theorem and others in detail in this article,