Question

A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. It makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Then n and Z for this atom will be (Ground state energy of hydrogen atom is – 13.6 eV)

- Z = 2
- Z = 4
- n = 1
- n = 2

Solution

The correct options are

**B** Z = 4

**D** n = 2

ΔE=204=13.6 Z2(11−14n2)

40.8=13.6Z2(1n2−14n2)

Solving we get,.

Z = 4 and n = 2

ΔE=204=13.6 Z2(11−14n2)

40.8=13.6Z2(1n2−14n2)

Solving we get,.

Z = 4 and n = 2

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