IB Physics SL E.1 Structure of the atom questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceEasy1 mark

A nucleus is represented by ZAX{}^{A}_{Z}X A second nucleus has the same proton number but one more neutron. What is its nuclear notation?

Show answer and explanation

Answer: A. AA is the nucleon number and ZZ is the proton number. One additional neutron increases the nucleon number by 1 while leaving the proton number, and therefore the element XX unchanged. The notation is ZA+1X{}^{A+1}_{Z}X

B: increases the proton number while keeping the nucleon number unchanged. C: decreases the nucleon number, corresponding to one fewer neutron at the same proton number. D: increases both the proton and nucleon numbers, adding a proton as well as a neutron.

Remember: adding a neutron changes AA not ZZ

Question 2

Paper 2Easy6 marks

Figure 1 shows nuclear notation and schematic emission and absorption spectra for an element.

2311Xemission spectrumabsorption spectrum400450500550wavelength / nm
(a)

In the nuclear notation shown in Figure 1, state what AA and ZZ represent, and calculate the number of neutrons in the nucleus.

[3]
(b)

Describe one feature of the line positions in the spectra shown in Figure 1.

[1]
(c)

Explain how the spectra provide evidence that atomic energy levels are discrete.

[2]

Question 3

Multiple choiceMedium1 mark

An atom emits a photon during the transition shown in the energy-level diagram. Take h=6.6×10−34 J sh=6.6\times10^{-34}\,\text{J s} and c=3.0×108 m s−1c=3.0\times10^8\,\text{m s}^{-1} What is the wavelength of the emitted photon?

Energy, E / J5.0×10−19 J4.0×10−19 J3.0×10−19 J2.0×10−19 J1.0×10−19 J0 J0 J1.1×10−19 J4.4×10−19 J
Show answer and explanation

Answer: C. The photon energy equals the difference between the atomic energy levels. Using E=hfE=hf and c=fλc=f\lambda λ=hc/ΔE\lambda=hc/\Delta E

  1. ΔE=(4.4−1.1)×10−19=3.3×10−19 J\Delta E=(4.4-1.1)\times10^{-19}=3.3\times10^{-19}\,\text{J}
  2. λ=(6.6×10−34)(3.0×108)3.3×10−19=6.0×10−7 m\lambda=\frac{(6.6\times10^{-34})(3.0\times10^8)}{3.3\times10^{-19}}=6.0\times10^{-7}\,\text{m}

A: adds the two level energies instead of finding their difference.
B: uses the upper level’s energy relative to zero rather than the energy gap shown.
D: uses the lower level’s energy as the photon energy rather than the difference between the levels.

Remember: a spectral line corresponds to a photon with energy equal to the difference between two discrete atomic energy levels.

The best questions are still waiting

Unlock every question in this topic, then type, photograph or draw your answer and have it marked against the markscheme. Find out where you are dropping marks.