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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2025

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In a hydrogen like ion, the energy difference between the $2^{nd}$ excitation energy state and ground is 108.8 eV. The atomic number of the ion is

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Answer:

Solution:

$Δ E = 13.6 z^{2} [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$
$(13.6) z^{2} [\frac{1}{1} - \frac{1}{9}] = 108.8$
$\frac{( 13.6 ) ( 8 )}{9} (z^{2}) = 108.8$
$z = 3$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2025

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