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प्रश्न
Answer the following question.
State Bohr's quantization condition of angular momentum. Calculate the shortest wavelength of the Bracket series and state to which part of the electromagnetic spectrum it belongs.
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उत्तर
According to Bohr's quantization, the electrons revolve around the nucleus only in those orbits for which the angular momentum is the integral multiple of
`h/(2pi)`
`L = (nh)/2`
for bracket series n2 = ∞
`1/λ = R_HZ^2{1/4^2 - 1/∞}`
`1/λ = R_H/16`
`λ = 16/R_H = 14.58 xx 10^-7`m
This wavelength belongs to the infrared region.
संबंधित प्रश्न
Define ionization energy.
Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the nth orbit and the particle’s speed v in the orbit depend on n as ______.
What does Bohr’s second postulate state about the electron orbiting around the nucleus?
Who explained the quantisation of the orbiting electron’s angular momentum in 1923?
What does the de Broglie hypothesis propose about material particles such as electrons?
Under what conditions can particle waves produce standing waves?
What happens when a string is plucked?
What does the condition \[2\pi r_n=n\lambda\] require of the orbit?
What is the electron’s de Broglie wavelength when its momentum magnitude is \[mv_n\]?
