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For the hydrogen like atom, the energy of the ground state is -54.4 eV. The value of $Z$ is

Answer:

Solution:

$E_{1} = - 13.6 Z^{2} eV$ . $13.6 Z^{2} = 54.4 \Rightarrow Z^{2} = 4 \Rightarrow Z = 2$ (singly ionised helium, $He^{+}$ ).

Subject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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