Question

Rubidium was named after the red color observed in a flame test based on its emission at 645.8 nm. What is the total energy released , in kilojoules, by this emission when 37.0g of rubidium is subjected to a flame test? Assume each rubidium atom emits one photon at 645.8 nm.

Answer 1

Energy released by one photon

E = h*c/wavelength

= 6.626*10^-34 * 3*10^8 / ( 645.8*10^-9)(in m)

= 3.08*10^-19 joules.

This energy is produced by 1 photon of 1 atom

Now moles of rubidium = mass / molar mass

= 37.0 / 85.5

= 0.4327 moles

1 moles have 6.022*10^23 atoms

0.4327 moles will have 6.022*10^23*0.4327

= 2.606*10^23 atoms

Total energy released by all the atoms

(2.606*10^23)*(3.08*10^-19)

= 80264.8 joules

Or 80.2648 kilojoules.

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