Question

Barium imparts a characteristic green color to a flame. The wavelength of this light is 551 nm. Determine the energy involved in kJ/mol.

A) 2.25 kJ/mol2.25kJ/mol
B) 3.50 kJ/mol3.50kJ/mol
C) 4.78 kJ/mol4.78kJ/mol
D) 6.20 kJ/mol6.20kJ/mol

Answer 1

The energy involved in the green color imparted to a flame by barium is 2.25 kJ/mol.

Step-by-step explanation:

The energy involved in the green color imparted to a flame by barium can be determined using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light. To convert the wavelength from nm to m, divide it by 10^9. Plugging in the values, we have E = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (551 x 10^-9 m) = 3.617 x 10^-19 J. To convert this energy to kJ/mol, divide it by Avogadro's constant (6.022 x 10^23), and then multiply by 10^-3. Therefore, the energy involved in kJ/mol is 3.617 x 10^-19 J / (6.022 x 10^23) x 10^-3 = 2.25 kJ/mol.