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ankoles [38]
3 years ago
6

How many degenerate orbitals can be found in each p subshell?

Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

3

Explanation:

For p-subshell we have the value of angular quantum number as l = 1 & the degenerate orbitals are numbered from -l to +l.

  • Therefore here we have -1,0,+1 as the three degenerate orbitals of p subshell.

This is in accordance with the Schrodinger's model that allows the electron to occupy a three dimensional space. It therefore requires three quantum numbers, to describe the orbitals in which electrons can be found. The three coordinates that come from Schrodinger's wave equations are the principal (n), angular (l), and magnetic (m) quantum numbers.

These quantum numbers decide the shape, size and orientation in space of the orbitals on an atom.

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Give two example for push or pull to change the state of motion of time two examples​
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Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of
Evgesh-ka [11]

Answer:

a)  V=7.5m/s

b) rms=8.4m/s

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

  Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\

Generally the equation for Average speed is mathematically given by

 V_{avg}=\frac{\sum(nv)}{N}

Therefore

 V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}

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b)

Generally the equation for RMS speed of the particle is mathematically given by

  rms=\sqrt{\frac{\sum(nv^2)}{N}}

  rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}

  rms=\sqrt{69.73}

  rms=8.4m/s

c

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Answer:

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Explanation:

because i calculated

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