An experiment looking at structures smaller than a cell would most likely employ a

Chemistry

1 answer:

Veseljchak [2.6K]2 years ago

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Answer:

In a conventional optical microscope, objects less than about 200 nanometers apart cannot be distinguished from one another. ... Although electron microscopes produce a detailed image of very small structures, they cannot provide an image of the proteins that make up those structures.

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Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
$m_l = 0$ .
Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

The quantum number $n$ (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, $n = 4$ for this electron.

The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

The spin quantum number $m_s$ distinguishes between the two electrons in an orbital. The two possible values of $m_s \!$ are $(+1/2)$ and $(-1/2)$ . Typically, the first electron in an orbital is assigned an upward ( $\uparrow$ ) spin, which corresponds to $m_s = (+1/2)$ .