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In a laboratory experiment, light from a hydrogen discharge lamp is used to produce a line emission spectrum - Scottish Highers Physics - Question 10 - 2018

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In a laboratory experiment, light from a hydrogen discharge lamp is used to produce a line emission spectrum. The line spectrum for hydrogen has four lines in the vi... show full transcript

Worked Solution & Example Answer:In a laboratory experiment, light from a hydrogen discharge lamp is used to produce a line emission spectrum - Scottish Highers Physics - Question 10 - 2018

Step 1

State two features of the Bohr model of the atom.

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Answer

  1. The Bohr model describes the atom as having a central positively charged nucleus.
  2. Negatively charged electrons orbit the nucleus in discrete energy levels or shells, without radiating energy.

Step 2

Determine the frequency of the photon emitted when an electron makes this transition.

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Answer

To find the frequency of the photon emitted during the transition from E1 to E3:

  1. Calculate the change in energy (ΔE): extΔE=E3E1=2.42×1019J(0.871×1019J)=1.55×1019J ext{ΔE} = E3 - E1 = -2.42 \times 10^{-19} \, \text{J} - (-0.871 \times 10^{-19} \, \text{J}) = -1.55 \times 10^{-19} \, \text{J}
  2. Use the equation relating energy and frequency: E=hff=EhE = hf \rightarrow f = \frac{E}{h} where h = Planck's constant, approximately 6.63×1034Js6.63 \times 10^{-34} \, \text{Js}.
  3. Substitute values: f=1.55×10196.63×10342.33×1014Hzf = \frac{1.55 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 2.33 \times 10^{14} \, \text{Hz}

Step 3

Determine the recessional velocity of the distant galaxy.

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Answer

To find the recessional velocity (v) of the distant galaxy, we use the formula:

  1. First, determine the shift in wavelength: z=λobservedλemittedλemitted=661 nm656 nm656 nm0.007628z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = \frac{661 \text{ nm} - 656 \text{ nm}}{656 \text{ nm}} \approx 0.007628
  2. Relate the redshift (z) to velocity using the formula: v=zcv = zc where c ≈ 3.00×108m/s3.00 \times 10^8 \, \text{m/s}.
  3. Substituting the values: v=0.007628×(3.00×108)2.29×106m/sv = 0.007628 \times (3.00 \times 10^8) \approx 2.29 \times 10^6 \, \text{m/s}

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