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VMariaS [17]
3 years ago
10

Arrange the following atoms according to decreasing effective nuclear charge experienced by their valence electrons: Cl, Na, P,

and S. Rank from largest to smallest. To rank items as equivalent, overlap them.
Cl Na P S
Largest Smallest
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

Cl < S < P < Na

Explanation:

We have to understand the trend for effective nuclear charge before answering the question;

The periodic table tendency for effective nuclear charge: Increase across a period (due to increasing nuclear charge with no accompanying increase in shielding effect). Decrease down a group (although nuclear charge increases down a group, shielding effect more than counters its effect).

Based on this, the elements can be arranged in the following order;

In decreasing order;

Cl < S < P < Na

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
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The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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