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)

• A. Z = 2
• B. Z = 4
• C. n = 1
• D. n = 2

Answer 1

Answer: (B), (D)

The correct options are
B Z = 4
D n = 2

$\Delta E=204=13.6Z^2(\frac{1}{1}-\frac{1}{4n^2})$

$40.8=13.6Z^2(\frac{1}{n^2}-\frac{1}{4n^2})$

Solving we get,.

Z = 4 and n = 2