Question

the argon - ion laser has two major emission lines, at 488 and 514 nm. each of these emissions leaves the ar⁺ ion in an energy level that is 2.76×10⁻¹⁸ j above the ground state. (a) calculate the energies of the two emission wavelengths in joules. e₄₈₈ = i×10 i j e₅₁₄ = i×10 i j etextbook and media

Explanation:

Step1: Recall energy - wavelength formula

The energy of a photon is given by $E = h\frac{c}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant), $c= 3\times 10^{8}\ m/s$ (speed of light) and $\lambda$ is the wavelength.

Step2: Calculate $E_{488}$

First, convert $\lambda_{488}=488\ nm = 488\times 10^{-9}\ m$ to SI units. Then, $E_{488}=h\frac{c}{\lambda_{488}}=(6.626\times 10^{-34}\ J\cdot s)\times\frac{3\times 10^{8}\ m/s}{488\times 10^{-9}\ m}\approx4.07\times 10^{-19}\ J$.

Step3: Calculate $E_{514}$

Convert $\lambda_{514}=514\ nm = 514\times 10^{-9}\ m$ to SI units. Then, $E_{514}=h\frac{c}{\lambda_{514}}=(6.626\times 10^{-34}\ J\cdot s)\times\frac{3\times 10^{8}\ m/s}{514\times 10^{-9}\ m}\approx3.87\times 10^{-19}\ J$.

Answer:

$E_{488}=4.07\times 10^{-19}\ J$
$E_{514}=3.87\times 10^{-19}\ J$