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.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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not clear