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garik1379 [7]
4 years ago
12

To show the electron configuration for an atom, when would it be better to use an orbital notation than to use a written configu

ration with numbers, letters, and superscripts
Physics
2 answers:
fgiga [73]4 years ago
8 0

Answer:

The correct answer is how the electron distributes itself in shells

Explanation:

An orbital documentation is progressively fitting on the off chance that you need to demonstrate how the electrons of an iota are disseminated in each subshell. This is on the grounds that there are a few iotas that have unique electronic designs that aren't clear in simply composed setups.

To demonstrate the electron setup for an atom, it would be better to utilize an orbital documentation than to utilize a composed design with numbers, letter and superscripts when the point is to indicate orbital shapes in subshells.

Clarification : When we expect to demonstrate the orbital shapes in sub shells, it implies that we are meaning to demonstrate the electron turns for that component or atom. The images used to speak to electrons are these ↑↓ bolts. At the point when both the bolt are inverse way that implies it drops the turn of one another and is in unbiased state.

lakkis [162]4 years ago
5 0

when the aim is to show electron distributions in shells.

hope this helps :)

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An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
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Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

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now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

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intensity is defined as

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now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

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I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

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i hope this if not sorry

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