Questions & Answers

Question

Answers

Answer

Verified

128.4k+ views

According to the De-Broglie hypothesis, every particle has a wavelength associated with it and relates the wavelength of that particle with its momentum. Before the 19th century, there was a huge debate on whether light propagates as a particle or a wave. There were many great scientists who favored the particle nature of light (example: Issac Newton) and many who favored wave nature (example: Young Thomas- Who stated after the result of his Young’s double slit experiment). After the Photoelectric effect, Einstein proved that light has particle nature and gave many supportive reasons for so. But the debate still remains as the particle nature of light couldn’t explain the interference of light which was the result of Young’s double slit experiment.

De-Broglie then gave his thesis in which he states that a light has both particles as well as wave nature.

Hence considering this postulate, he derived the relation between momentum and wavelength of light. Einstein suggested that the light is made up of small packets called quanta or photons. Hence if light is a particle, it must have energy associated with it, given by:

$E=mc^2$ [ Einstein’s special theory of relativity ]

And if light is wave, its energy is given by Planck's theory, given by:

$E=\dfrac{hc}{\lambda}$

As both energy are the same, hence equating:

$mc^2=\dfrac{hc}{\lambda}$

Or $\lambda = \dfrac{h}{mc}$

And $m\times c= p, i.e\ momentum\ of\ particle$.

Hence $\lambda = \dfrac{h}{p}$

Now, in his hypothesis, he suggested that it’s not valid only for light but valid for any particle as small or as large as possible.

Experiments have shown that the lighter particles moving with some velocities possess their unique wavelength associated with them.

It’s true that every object like a car moving on road and man running on track possess their own wavelengths. But the wavelengths are so small to observe that naked eye can’t detect them. This is because in expression $\lambda = \dfrac{h}{p}$, Planck's constant (h) = $6.62\times 10^{-34} m^2 kg s^{-1}$is so small that for heavier objects, wavelength tends to zero and only lighter particles like electrons ($m_e =9.1\times 10^{-31}kg$) can show the wavelength.