
How did the Broglie hypothesis lead to Bohr’s quantum condition of atomic orbits?
Answer
506.4k+ views
Hint: De Broglie's hypothesis of matter waves postulates that any particle of matter that has linear momentum is also a wave. The wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The speed of the matter wave is the speed of the particle.
Complete answer:
According to Bohr's Quantum Condition “Only those atomic orbits are allowed as stationary orbits in which angular momentum of electrons is an integral multiple of $\dfrac{h}{{2\pi }}$"
If m is the mass, v is the velocity and r is the radius of the orbit, then angular momentum of electron,\[L = mvr\]
According to Bohr's Quantum Condition,
$mvr = \dfrac{h}{{2\pi }}$ ...(i)
According to De Broglie Quantum condition only those atomic orbits are allowed as stationary orbits in which circumference of electron orbit is integral multiple of de-broglie wavelength associated with electron i.e. $2\pi r = n\lambda $ ...(ii)
According to de broglie hypothesis,
$\lambda = \dfrac{h}{{mv}}$ ...(iii)
Substituting this value in (ii) we get,
$2\pi r = n\left( {\dfrac{h}{{mv}}} \right)$
$ \Rightarrow mvr = \dfrac{{nh}}{{2\pi }}$
This is Bohr's Quantum condition
Note: Limitation of the de-broglie's wavelength is only applicable to microscopic particles such as protons, neutrons, electrons etc. It is not applicable for macroscopic objects like cricket ball, football etc. These objects have wavelengths but it is so small that it does not have any real existence.
Complete answer:
According to Bohr's Quantum Condition “Only those atomic orbits are allowed as stationary orbits in which angular momentum of electrons is an integral multiple of $\dfrac{h}{{2\pi }}$"
If m is the mass, v is the velocity and r is the radius of the orbit, then angular momentum of electron,\[L = mvr\]
According to Bohr's Quantum Condition,
$mvr = \dfrac{h}{{2\pi }}$ ...(i)
According to De Broglie Quantum condition only those atomic orbits are allowed as stationary orbits in which circumference of electron orbit is integral multiple of de-broglie wavelength associated with electron i.e. $2\pi r = n\lambda $ ...(ii)
According to de broglie hypothesis,
$\lambda = \dfrac{h}{{mv}}$ ...(iii)
Substituting this value in (ii) we get,
$2\pi r = n\left( {\dfrac{h}{{mv}}} \right)$
$ \Rightarrow mvr = \dfrac{{nh}}{{2\pi }}$
This is Bohr's Quantum condition
Note: Limitation of the de-broglie's wavelength is only applicable to microscopic particles such as protons, neutrons, electrons etc. It is not applicable for macroscopic objects like cricket ball, football etc. These objects have wavelengths but it is so small that it does not have any real existence.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

