Acetaldehyde shows two UV bands, one with a lmax of 289 nm ( 5 12) and one with a lmax of 182 nm ( 5 10,000). Which is the n p* transition and which is the p p* transition? Explain your reasoning.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

Explanation:

The two types of acetaldehyde transition are as follows:

n→π* and π→π*

From the attached diagram we have to:

ΔEn→π* < ΔEπ→π*

ΔEα(1/λ)

Thus:

λn→π* > λπ→π*

In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.

The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied