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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

PhysicsEasysingle choice

As per Bohr model, the minimum energy (in $eV$ ) required to remove an electron from the ground state of doubly ionized Li atom ( $Z = 3$ ) is

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Answer:

Solution:

For hydrogen and hydrogen like atoms
$E_{n} = - 13.6 \frac{(Z^{2})}{(n^{2})} eV$
Therefore,ground state energy of doubly ionized lithium atom (Z=3,n=1) will be

$E_{n} = - 13 \cdot 6 \frac{{(3)}^{2}}{\frac{{(3)}^{2}}{{(1)}^{2}}} = - 122 \cdot 4 e V$
$∴$ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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