Question

A hydrogen like atom (atomic number Z) is in a higher excited state of quantam number n. This excited atom can make a transition to the first excited state by emitting a photon of energy 27.2eV. Alternatively the atom from the same excited state can make a transition to second excited state by emitting a photon of energy 10.20 eV The value of n and z are given (ionization energy of hydrogen atom is 13.6eV)

• A. n = 6 and z=3
• B. n = 3 and z = 6
• C. n = 8 and z = 4
• D. n= 4 and z = 8

Answer 1

Answer: (A)

n = 6 and z=3

Answer 2

$\Delta E_1 = 27.2 = (13.6) \left[ \frac{1}{4} - \frac{1}{n^2} \right] z^2$ $\Delta E_2 = 10.2 = (13.6) \left[ \frac{1}{9} - \frac{1}{n^2} \right] z^2$