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Anna007 [38]
3 years ago
13

Whiat is quantum numbers describes the size and energy of an orbital?

Physics
1 answer:
Levart [38]3 years ago
7 0

Answer:

The answer is the principal Quantum number (n)

Explanation:

The principal quantum number is one of the four quantum numbers associated with an atom.

It is denoted by a number n=1,2,3,4 etc

It tells both size (directly) and energy (indirectly) of an orbital.

When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.

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The number of protons
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14-6=8
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3 years ago
Which of the following correctly describe electric field lines?
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-- Electric field lines DO never cross.  <em>(A) </em>

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What is a pure substances​
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a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}

Now, we have that

{v_f}^2-{v_0}^2=2a\Delta x

so we end up with a distance traveled of

\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=79.6\,\mathrm m

6 0
3 years ago
An object is thrown directly upwards from the ground at a velocity of 9ms. Recalling that the acceleration due to gravity is −gm
olga55 [171]

Answer:

Explanation:

Given

Object is thrown with a velocity of u=9\ m/s

Acceleration due to gravity is -g (i.e. acting downward)

Vertical distance traveled by object is given by

v^2-u^2=2as  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

at maximum height final velocity is zero

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s=\frac{81}{2g}=\frac{40.5}{g}\ m

time taken to reach maximum height

using

v=u+at

0=9-gt

t=\frac{9}{g}\ s

6 0
4 years ago
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