Additional Answer Options: 11.3

2018 - November - Paper 1 - Question 11

11.1

The work function of a metal is the minimum energy needed to eject an electron from the metal surface.checkcheck

2 marks

(1 x 2 marks)

11.2

 

Potassium/Kcheck 

fo for potassium is greater than fo for caesiumcheck

Or

Work function is directly proportional to threshold frequency.check

light bulb Tips: 

  • The work function of the metals is calculated using W0 = hf0, therefore, the metals with the higher threshold frequency will have a higher work function.
  • Potassium has a higher work function.

2 marks

(2 x 1 mark)

11.3

Option 1

c=f\lambdacheck

3\times10^8=f\left(5,5\times10^{-7}\right)check

f=5,45\times10^{14}\mathrm{~Hz}

\mathrm{f}_{\mathrm{uv}}<\mathrm{f}_{\mathrm{o}\text{ of }\mathrm{K}\text{ \lparen potassium\rparen }}

\therefore\text{ Ammeter in circuit B will not show a reading }check

light bulb Tips: 

  • For the photoelectric effect to occur, the frequency of the incident light must be higher than the threshold frequency.
  • You need to use the wave equation to determine the frequency of the photon that corresponds with the wavelength of the light.

There are more ways to answer this question. To view other options, click here.

3 marks

(3 x 1 mark)

11.4 

Option 1

\left.\begin{array}{l}E=W_0+E_{k(\max)}\\ hf=hf_{o}+\frac12mv_{\max}^2\\ h\frac{c}{\lambda}=h\frac{c}{\lambda_0}+E_{k(\max)}\end{array}\right\}Any onecheck

\frac{\left(6,63\times10^{-34}\right)\left(3\times10^8\right)^{}}{5,5\times10^{-7}}=\left(6,63\times10^{-34}\right)\left(5,07\times10^{14}\right)+E_{k(\max)}checkcheckcheck

Ek = 2,55 x 10-20 J (Range 2,52 x 10-20 - 2,6 x 10-20 J) check

Or

Option 2

\left.\begin{array}{l}E=W_0+E_{k(\max)}\\ hf=hf_{o}+\frac12mv_{\max}^2\\ h\frac{c}{\lambda}=h\frac{c}{\lambda_0}+E_{K(\max)}\end{array}\right\}Any onecheck

\left(6,63\times10^{-34}\right)\left(5,45\times10^{14}\right)=\left(6,63\times10^{-34}\right)\left(5,07\times10^{14}\right)+E_{k(\max)}checkcheckcheck

E_{K}=2,52\times10^{-20}\mathrm{J}\quad\left(\text{ Range: }2,52\times10^{-20}-2,6\times10^{-20}\mathrm{J}\right)check

light bulb Tip: you can work out each of the parts separately and later put them into the equation. 

5 marks

(5 x 1 mark)

11.5

 

Remains the same check

light bulb Tips:

  • The intensity of the light affects the number of photons striking the metal per unit of time.
  • The energy of the photon depends only on the frequency of the light.
  • The frequency of the light does not change when the intensity increases.
  • All of the energy of the photon is passed on to the electron that is released, therefore, the energy of the electron remains the same.

1 mark

 

Or

Option 2

\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{\left(6,63\times10^{-34}\right)\left(3\times10^8\right)}{5,5\times10^{-7}}

=3,6164\times10^{-19}\mathrm{~J}

\mathrm{W}_0=hf_0=\left(6,63\times10^{-34}\right)\left(5,55\times10^{14}\right)=3,68\times10^{-19}\mathrm{~J}

\mathrm{W}_0>\mathrm{E}\text{ or }\mathrm{hf}_0>\mathrm{hf}

\therefore\text{ The ammeter will not register a current }

Or

Option 3

c=f_0\lambda_0

3\times10^8=\left(5,55\times10^{14}\right)\lambda

\lambda_0=5,41\times10^{-7}\mathrm{~m}

\lambda_{\circ}\text{ \lparen threshold wavelength\rparen }<\lambda\text{ \lparen incident wavelength\rparen }

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