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PhysicsModern PhysicsAtomic structureEasy2 minQB

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A $H e^{+}$ ion is in its first excited state. Its ionization energy is:

Answer:

Solution:

$E = \frac{- ( 13.6 ) z ^{2}}{n ^{2}}$

$= \frac{- ( 13.6 ) ( 2 ) ^{2}}{2 ^{2}} = - 13.60 e V$

Total energy of $H e^{+}$ ion is in its first excited state is $- 13.60 e V$ . Hence, its ionization energy will be $13.60 e V$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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