Answer to Question #87303 in General Chemistry for Sampson

Question

A 3.8mg sample organic compound gave 5.80mg co2 and 1.58H2o on combustion.
Calculate the energy frequency and wave length (in nm)associated with transition of electron from N=3 to N=1

Answer 1

n(CO₂) = m/M = 5.8g/44g/mol = 0.1318 mol

n(H₂O) = 1.58g/18g/mol = 0.0878 mol

n(CO₂) : n(H₂O) = 0.1318 : 0.0878 = 1.5 : 1 = 3 : 2

So the formula for this organic compound is C₃H₄ .

I assume you are talking about the transition of an electron in hydrogen. Any transition to energy level 1 will result in the emission of UV light. 

Start by computing the energy of the electron at n=3 

E(n) = -Rh(1/n²) where Rh (Rydberg constant) = 2.18x10^-18 J 

E(3) = -Rh(1/9) 

E(1) = -Rh 

Get the difference in energy, DE = E(1) - E(3) 

Then compute the wavelength from the difference in energy: 

E = hc/lambda, where h = Planck's constant, c = speed of light, lambda is the wavelength in meters. 

lambda = hc/DE 

When you've done all this, you should get 1.03 x 10^-7 m or 103 nm.