When light is shined on a thin sheet of metal, electrons are released with a kinetic energy of 6.81 x 10-21 J. If the binding energy of themetal is known to be 6.25 x 10-19 J, what is the wavelength of the light, in nm?Do not report units in your answer. Report your answer to one decimal places.

Answer:- This question is answered by using the simple concept of photoelectric effect in which…

Answered: When light is shined on a thin sheet of…

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter25: Nuclear Chemistry

Section: Chapter Questions

Problem 62GQ

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Similar questions

True or false? Rutherford’s bombardment experiments with metal foil suggested that the particles were being deflected by coming near a large, positively charged atomic nucleus.
A B58 atom (mass = 8.0246 amu) decays into B48 atom (mass = 8.0053 amu) by loss of a + panrticle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?
Slaters rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the shielding constant, S, is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z. (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) Z = Z S The shielding constant, S, is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1s 0.30 works better). b) Electrons in n 1 groups contribute 0.85. c) Electrons in n 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00. As an example, let us calculate Z for the outermost electron of oxygen: S = (2 0.85) + (5 0.35) = 3.45 Z = 8 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z = 28 [18 1.00] [7 0.35] = 7.55 and for an s electron in Ni: Z = 28 [10 1.00] [16 0.85] [1 0.35] = 4.05 (Here 3s, 3p, and 3d electrons are in the (n 1) groups.) a) Calculate Z for F and Ne. Relate the Z values for O, F, and Ne to their relative atomic radii and ionization energies. b) Calculate Z for one of the 3d electrons of Mn, and compare this with Z for one of the 4s electrons of the element. Do the Z values give us some insight into the ionization of Mn to give the cation?

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When light is shined on a thin sheet of metal, electrons are released with a kinetic energy of 6.81 x 10-21 J. If the binding energy of the metal is known to be 6.25 x 10-19 J, what is the wavelength of the light, in nm? Do not report units in your answer. Report your answer to one decimal places.