The work function of palladium (Pd) is 5.22 eV. What is the minimum frequency of light to observe the photoelectric effect from
a Pd surface? If light with a 200 nm wavelength is absorbed by the Pd surface, what is the velocity of the emitted electrons?
1 answer:
Answer:
1)4×10^13Hz
2) 9.95×10-9J
Explanation:
From the image attached, it is clear that the work function of the palladium metal must first be obtained in joules. Then, the frequency is obtained from E=hf.
The kinetic energy if the photoelectrons is obtained as the difference between the energy of the photon and the work function of the metal.
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Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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