A doubly-ionized lithium atom $\mathrm { Li } ^ { + + }$, which has 3 protons, undergoes a transition from the $n = 5$ state to the $n = 4$ state. What is the wavelength of the photon which is emitted during this
transition? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} \right)$
A) $650 \mathrm {~nm}$
B) $350 \mathrm {~nm}$
C) $550 \mathrm {~nm}$
D) $750 \mathrm {~nm}$
E) $450 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The energy difference between the levels can be calculated using the formula for hydrogen-like atoms: $E = -13.6 \times Z^2 \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV, where $Z$ is the atomic number (3 for lithium). The energy difference is then converted to wavelength using $\lambda = \frac{hc}{E}$. For the transition from $n=5$ to $n=4$, the wavelength is approximately 450 nm.

A Doubly-Ionized Lithium Atom Li++\mathrm { Li } ^ { + + }Li++

A Doubly-Ionized Lithium Atom $\mathrm { Li } ^ { + + }$