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PhysicsModern PhysicsAtomic structureHard2 minPYQ_2014

PhysicsHardsingle choice

A triply ionized beryllium $(Be^{3 +})$ has the same orbital radius as the ground state of hydrogen. Then the quantum state $n$ of $Be^{3 +}$ is

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Radius of $n^{th}$ orbit in hydrogen like atoms is $r_{n} = \frac{a _{0} n ^{2}}{Z}$ where $a_{0}$ is the Bohr's radius For hydrogen atom, $Z = 1$ $∴ r_{1} = a_{0} (∵ n = 1$ for ground state $)$ For $Be^{3 +}, Z = 4$ $∴ r_{n} = \frac{a _{0} n ^{2}}{4}$ According to given problem $r_{1} = r_{n}$ $a_{0} = \frac{n ^{2} a _{0}}{4}$ $n = 2$

Stream:AIIMSSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2014

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