Which of the following statements about ions is NOT CORRECT?
1 answer:
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Answer:
a) v₀ = 4.41 × 10¹⁴ s⁻¹
b) W₀ = 176 KJ/mol of ejected electrons
c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.
d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.
e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.
Explanation:
The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.
a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m
Speed of light = 3 × 10⁸ m/s
The frequency of the light, v₀ = ?
Frequency = speed of light/wavelength
v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹
b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀
h = Planck's constant = 6.63 × 10⁻³⁴ J.s
E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J
E in J/mol of ejected electrons
Ecalculated × Avogadros constant
= 2.92 × 10⁻¹⁹ × 6.023 × 10²³
= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons
c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).
As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.
d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.
e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,
And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s
Hope this Helps!!!
Answer:
Z* = 3.55
Explanation:
Slater rule says that:
Z*= Z - S
Z* be the nuclear effective charge
Z is the nuclear charge
S is the shielding constant
First we write the electronic configuration of platinum:
The first Slater rule says that we need to group:
The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.
For the third rule we have two considerations, if is ns or np and if is nd or nf:
For our case, we have an electro that is in ns, so the rule says that
-electrons within same group shield 0.35, except the 1s which shield 0.30
-electrons within the n-1 group shield 0.85
-electrons within the n-2 or lower groups shield 1.00
Now we can proceed with the calculation:
The first consideration in the third rule does not apply as we only have one electron on this level.
The second consideration will be as follow for the level 5, where we have 17 electrons.
Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.
So the result for S=(60*1.00 + 17*0.85) = 74.45
And the equation is: Z* = 78 - 74.45
So Z* = 3.55