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Chem Studio by Learn4Less

Element 78

Platinum (Pt)

78
Pt

platinum

Metal · d block · group 10, period 6

Electron configuration
[Xe] 6s1 4f14 5d9[\mathrm{Xe}]\,6s^{1}\,4f^{14}\,5d^{9}
Unpaired electrons
2 (paramagnetic)
Molar mass
195.08 g/mol
Electronegativity
2.28
Atomic radius
139 pm
First ionization energy
870 kJ/mol
Next ionization energies
1,791
Electron affinity
205 kJ/mol released

Practice

Practice platinum

Questions from the course about platinum. If you’re signed in, your answers count toward your progress.

  1. Question 1Photoelectric effect: kinetic energy of ejected electronsExam level

    Light of wavelength 205 nm205\ \mathrm{nm} shines on a clean platinum surface. The work function of platinum is ϕ=5.65 eV\phi = 5.65\ \mathrm{eV} (1 eV=1.602×10−19 J1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}). What is the maximum kinetic energy of the ejected electrons, in J?

    J\mathrm{J}

    Scientific notation: type 4.86e-7 or 4.86 x 10^-7.

  2. Question 2Photoelectric threshold and work functionExam level

    The work function of platinum is ϕ=5.65 eV\phi = 5.65\ \mathrm{eV} (1 eV=1.602×10−19 J1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}). What is the threshold frequency, in s−1\mathrm{s^{-1}}, for the photoelectric effect in platinum?

    s−1\mathrm{s^{-1}}

    Scientific notation: type 4.86e-7 or 4.86 x 10^-7.

  3. Question 3Will these photons eject electrons?Exam level

    The work function of platinum is ϕ=5.65 eV\phi = 5.65\ \mathrm{eV} (1 eV=1.602×10−19 J1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}). Which of these wavelengths of light will eject electrons from platinum? Select all that apply.