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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, with electrons orbiting in discrete energy levels or shells.

  2. Electrons can only occupy specific energy levels, and transitions between these levels result in the emission or absorption of photons, which generates the line spectrum.

Step 2

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

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Answer

To find the frequency, we first calculate the energy difference between the two levels:

DeltaE=E2−E4=(−1.36×10−19 extJ)−(−5.45×10−19 extJ) \\Delta E = E₂ - E₄ = (-1.36 \times 10^{-19} \, ext{J}) - (-5.45 \times 10^{-19} \, ext{J})

DeltaE=4.09×10−19 extJ \\Delta E = 4.09 \times 10^{-19} \, ext{J}

Using the relationship between energy and frequency: E=hfE = h f where h is Planck's constant (approximately 6.63×10−34 J s6.63 \times 10^{-34} \, \text{J s}), we find:

f=ΔEh=4.09×10−196.63×10−34≈6.17×1014 Hzf = \frac{\Delta E}{h} = \frac{4.09 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 6.17 \times 10^{14} \, \text{Hz}

Step 3

Determine the recessional velocity of the distant galaxy.

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Answer

Using the formula for redshift: z=λobserved−λemittedλemittedz = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}

For the wavelengths given, z=661 nm−656 nm656 nm=5656≈0.00763z = \frac{661 \text{ nm} - 656 \text{ nm}}{656 \text{ nm}} = \frac{5}{656} \approx 0.00763

The recessional velocity can be calculated using: v=zcv = zc where c is the speed of light (3.00×108 m/s3.00 \times 10^8 \, \text{m/s}):

v=0.00763×3.00×108≈2.29×106 m/sv = 0.00763 \times 3.00 \times 10^8 \approx 2.29 \times 10^6 \, \text{m/s}

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