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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2009

PhysicsEasysingle choice

For the Bohr's first orbit of circumference $2 π$ , the de-Broglie wavelength of revolving electron will be

Answer:

Solution:

$m v r = \frac{nh}{2 π}$ , according to Bohr's theory $\Rightarrow 2 π r = n (\frac{h}{m v}) = nλ for n = 1, λ = 2 π r$

Stream:MHT_CETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2009

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