Speed of an electron in Bohr’s 7 t h orbit for Hydrogen atom is 3.6 × 1 0 6 m s − 1 . The corresponding speed of the electron in 3 r d orbit, in m s − 1 is :

For hydrogen like atom, the velocity of electron in n t h orbit is given by v = 2 ϵ 0 hn z 2 e 2 ∴ v ∝ n 1 ⇒ v 2 v 1 = n 1 n 2 ⇒ v 2 3.6 × 1 0 6 = 7 3 v 2 = 8.4 × 1 0 6 m s − 1

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PhysicsModern PhysicsAtomic structureEasy2 minQB

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Speed of an electron in Bohr’s $7^{t h}$ orbit for Hydrogen atom is $3.6 \times 1 0^{6} m s^{- 1}$ . The corresponding speed of the electron in $3^{r d}$ orbit, in $m s^{- 1}$ is :

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Answer:

Solution:

For hydrogen like atom, the velocity of electron in $n^{t h}$ orbit is given by

$v = \frac{z ^{2} e ^{2}}{2 ϵ _{0} hn}$

$∴ v \propto \frac{1}{n}$

$\Rightarrow \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}}$

$\Rightarrow \frac{3.6 \times 1 0 ^{6}}{v _{2}} = \frac{3}{7}$

$v_{2} = 8.4 \times 1 0^{6} m s^{- 1}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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