The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this problem. Part A What minimum frequency of light is required to ionize boron

The frequency is  

Explanation:

From the question we are told that

   The energy required to ionize boron is

Generally the ionization energy of boron pre atom is mathematically represented as

Here   is the Avogadro's constant with value

So

=>    

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

=>    

Here h is the Planks constant with value

So

=>      

(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