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10.1 Two small spheres, A and B, made of pure zinc are at rest 0.1 m apart on a wooden table - NSC Physical Sciences - Question 10 - 2023 - Paper 1

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10.1 Two small spheres, A and B, made of pure zinc are at rest 0.1 m apart on a wooden table. Sphere A is negatively charged and is free to move on the table, while ... show full transcript

Worked Solution & Example Answer:10.1 Two small spheres, A and B, made of pure zinc are at rest 0.1 m apart on a wooden table - NSC Physical Sciences - Question 10 - 2023 - Paper 1

Step 1

10.1.1 Define the term work function of a metal.

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Answer

The work function of a metal is defined as the minimum energy required to remove an electron from the surface of the metal. It represents the threshold energy that must be overcome for photoemission to occur.

Step 2

10.1.2 Explain, using a suitable calculation, why the ultraviolet light shone on sphere B will eject electrons from its surface.

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Answer

To determine if the ultraviolet light will eject electrons from sphere B, we can use the photoelectric equation:

E=hfE = hf

Where:

  • EE is the energy of the photons
  • hh is Planck’s constant (6.63×10346.63 × 10^{-34} Js)
  • ff is the frequency of the ultraviolet light (2.8×10162.8 × 10^{16} Hz)

Calculating the energy:

E=(6.63×1034extJs)(2.8×1016extHz)=1.86×1017extJE = (6.63 × 10^{-34} ext{ Js})(2.8 × 10^{16} ext{ Hz}) = 1.86 × 10^{-17} ext{ J}

Since the energy of the ultraviolet light (1.86 × 10^-17 J) is greater than the work function of zinc (6.63 × 10^-19 J), it will be sufficient to eject electrons from sphere B.

Step 3

10.1.3 Calculate the minimum number of photons of ultraviolet light that must strike sphere B which will cause sphere A to move from its rest position.

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Answer

To calculate the minimum number of photons required to exert a force on sphere A, we first determine the energy needed to overcome the force:

The charge on sphere A is q=5.4×106Cq = -5.4 × 10^{-6} C.

Using the force equation, F = rac{k imes |q_1| imes |q_2|}{r^2} we can calculate.

Rearranging gives us:

E=FimesdE = F imes d

Where:

  • F=0.027extNF = 0.027 ext{ N}
  • d=0.1extmd = 0.1 ext{ m}

Thus, the energy:

E=0.027extNimes0.1extm=0.00027extJE = 0.027 ext{ N} imes 0.1 ext{ m} = 0.00027 ext{ J}

Now, the energy per photon is Ephoton=1.86×1017extJE_{photon} = 1.86 × 10^{-17} ext{ J}.

Therefore, the minimum number of photons, nn, is:

n = rac{E}{E_{photon}} = rac{0.00027 ext{ J}}{1.86 × 10^{-17} ext{ J}}

Calculating gives:

ightarrow 1.45 × 10^{13} ext{ photons}$$

Step 4

10.2.1 Name the type of line spectrum observed.

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Answer

The type of line spectrum observed is an absorption spectrum.

Step 5

10.2.2 Describe the spectrum referred to in QUESTION 10.2.1.

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Answer

The absorption spectrum consists of a continuous spectrum of light with dark lines superimposed on it. These dark lines correspond to specific wavelengths of light that have been absorbed by the gas, indicating the presence of particular elements.

Step 6

10.2.3 Which one of the diagrams could result in the type of spectrum observed in QUESTION 10.2.1? Choose from DIAGRAM A or DIAGRAM B.

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Answer

The type of spectrum observed is consistent with DIAGRAM A, as it shows energy transitions that would lead to absorption of specific wavelengths of light.

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