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PhysicsModern PhysicsPhotoelectric effectMedium2 minQB

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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

Answer:

Solution:

$E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} eV$

For $Li^{2 +}$

$Z = 3$

and

$n = 1$

Ionization needs

$13.6 (3)^{2} = 122.4 eV$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: QB

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