The six metals have the work functions, W.

Part A Rank these metals on the basis of their cutoff frequency. Rank from largest to smallest. To rank items as equivalent, overlap them.

Part B Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface. Rank from largest to smallest. To rank items as equivalent, overlap them.

Part 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. (If no electrons are emitted from a metal, the maximum kinetic energy is zero, so rank that metal as smallest.) Rank from largest to smallest. To rank items as equivalent, overlap them.

Cesium= w= 2.1 eV Aliminium= w= 4.1 eV Beryllium= 5.0 eV Potassium= 2.3 eV Platinium= w= 6.4 eV Magnisium=w= 3.7 eV

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Vespertilio Vespertilio · Answer 1 · 2020-03-24T14:02:14+01:00

Answer:

Explanation:

W (Ce) = 2.1 eV

W (Al) = 4.1 eV

W (Be) = 5 eV

W (K) = 2.3 eV

W (Pt) = 6.4 eV

W (Mg) = 3.7 eV

Part A:

Work function is directly proportional to the cut off frequency.

let f denotes the cut off frequency.

So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)

Part B:

Maximum wavelength for the emission is inversely proportional to the cut off frequency

So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)

Part C:

E = 3.10 eV

Let K is the maximum kinetic energy

K = E - W

K (Ce) = 3.1 - 2.1 = 1 eV

K (Al) 3.1 - 4.1 = not possible

K (Be) = 3.1 - 5 = not possible

K (K) = 3.1 - 2.3 = 0.8 eV

K (Pt) = .1 - 6.4 = not possible

K (Mg) = 3.1 - 3.7 = not possible

So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)