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PhysicsModern PhysicsNuclear physicsEasy2 minPYQ_2023

PhysicsEasynumerical

If the binding energy of ground state electron in a hydrogen atom is $13.6 eV$ , then, the energy required to remove the electron from the second excited state of $Li^{2 +}$ will be: $x \times 10^{- 1} eV$ . The value of $x$ is _____.

Answer:

136.00

Solution:

Given:

Binding energy, $E_{H} = E_{0} = 13.6 eV$ .

Now for hydrogen like atom, we can write

$E_{{Li}^{2 +}} = E_{0} \frac{Z^{2}}{n^{2}} = 13.6 \times \frac{9}{9} = 13.6 eV$

$= 136 \times 10^{- 1} eV$

Therefore, $x = 136$ .

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2023

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