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PhysicsModern PhysicsAtomic structureMedium2 minQB

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The acceleration of an electron in the first orbit of the hydrogen atom ( $n = 1$ ) is :

Answer:

Solution:

For $H -$ atom $(n = 1)$

$\frac{m v ^{2}}{r} = \frac{1}{4 π ϵ _{o}} \frac{e ^{2}}{r ^{2}} ... (1)$

$m v r = \frac{h}{2 π} ... (2)$

On squaring $(2)$

$m^{2} v^{2} r^{2} = \frac{h ^{2}}{4 π ^{2}}$

On dividing both sides by $r^{3},$

$\frac{m ^{2} v ^{2}}{r} = \frac{h ^{2}}{4 π r ^{3}} \Rightarrow \frac{v ^{2}}{r} = \frac{h ^{2}}{4 π ^{2} r ^{3} m ^{2}}$

Acceleration $= \frac{h ^{2}}{4 π ^{2} r ^{3} m ^{2}}$ .

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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