The six metals have the work functions, E0.A. Rank these metals on the basis of their cutoff frequency. B. Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface. C. Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted electrons.1. Cesium: w = 2.1 eV 2. Aliminium: w = 4.1 eV 3. Beryllium: w = 5.0 eV 4. Potassium: w = 2.3 eV 5. Platinium: w = 6.4 eV 6. Magnesium: w = 3.7 eV

a) Platinum > Beryllium > Aluminum > Magnesium > Potassium > Cesium

b) Cesium > Potassium > Magnesium > Aluminum > Beryllium > Platinum

c) Cesium > Potassium

Explanation:

We must recall that the frequency of an electromagnetic wave is directly proportional to its energy. Hence as the work function of the metal increases, the minimum frequency required for emission of electrons occur increases accordingly.

Similarly, the maximum wavelength required for electron emission to occur varies inversely as the work function of the metal hence the answer provided.

Lastly, only caesium and potassium has work function less than the energy of the incident photon hence only these two metals experience electron emission with the kinetic energy of electrons emitted from caesium greater than that emitted from potassium.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.