Answer:
K.E(K) > K.E(Cs) > 0 (others)
Explanation:
Given the Work functions of the metal as
Aluminium (Wo)=4eV
Platinum(Wo) =6.4eV
Cesium (Wo) =2.1eV
Beryllium (Wo) = 5.0eV
Magnesium (Wo) = 3.7eV
Potassium (Wo) = 2.3eV
Using the formula:
K.E = hf - Wo........(1)
Wo = hfo..............(2)
From these the fo can be calculated for all the metals
Where K.E =Kinetic Energy
hf = energy of illumination = 3.10eV
h is Planck constant and has the value 6.6 × 10^-34JS^-1
The frequency f of the illumination is given by
f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34
f = 7.51 × 10¹⁴ Hz..........(*)
Now an electron is only ejected if the threshold frequency of the metal is reached.
The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.
We need to compare f with fo
If fo >= f there is emission, otherwise there is no emission
So using (2) we calculate for all fo and compare with f
K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)
K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)
K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)
K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)
K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)
K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)
So the metals whose electron gain Kinetic energy are:
Cesium
Potassium
Others have zero kinetic energy since no electron is emitted.
Hence the rank is:
K.E(K) > K.E(Cs) > 0 (others)
