Question

A hydrogen-like atom with the atomic number $Z$ is in the higher excited state of the quantum number ${}^{\prime }{n}^{\prime }$. This excited state atom can make a transition to the first excited state by successively emitted two photons of energy 10.2 eV and 17eV respectively.

Alternatively, the atom from the same excited state can make a transition to the 2nd excited state by emitting photons of energy 4.25 eV and 5.95 eV, respectively. Determine $Z$ and $n$?

• A. $n=6,z=3$
• B. $n=4,z=3$
• C. $n=2,z=2$
• D. $n=3,z=6$

Answer 1

Answer: (A)

$n=6,z=3$

According to the given condition,

The electron makes a transition from state $n_2$ is the first excited state releasing $10.2+17=27.2eV$10.2=17=27.2eV energy

$27.2eV=13.6{\left(z\right)}^2[\frac{1}{{\left(2\right)}^2}-\frac{1}{n^2}]\Rightarrow 2=z^2[\frac{1}{4}-\frac{1}{n^2}]$ ------ (1)

When the electron comes to the second excited state =4.25+5.95=10.2eV

(n1=the energy is 4.2+5.95=10.2eV 4.25=5.95=10.2e

$10.2eV=13.6\left(z^2\right)[\frac{1}{{\left(3\right)}^2}-\frac{1}{n^2}]\Rightarrow 0.75=2^2[\frac{1}{9}-\frac{1}{n^2}]$ ------- (2)

On solving we get, $z=3,n=6$.