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3A 4A 5A 6A 7A He BCN O F Ne Na Mg 3B 4B 5B 6B7BBB1B 28 Al Si P S Cl Ar sctv Cr Mn Fe Co Ni Cu Zn Ga Ge As Se srv2r Nb Mo Tc

ЗА 4А 5А 6А 74 Не BCNO Li Be Na Mg 3B 4B 5B 6B7BB 1B 2B Al Si P S Kca se tiver Mn Fe Co Ni Cu Zn Ga Rusly | 2 N Mac Ru Bhd CS
Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the be
Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the be
Use the References to access important values if needed for this question. Compare the de Broglie wavelength of a baseball mo
Compare the de Broglie wavelength of a bullet moving at 700 miles per hour (313 m/s) to that of an alpha particle moving at 3
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Answer #1

ANSWER:

1) THE ELEMENT IS Fe is because the last electron is filled in d-orbital and atomic number is 26

2)-the element is Si because the last electron is in p-orbital and the atomic number is 14.

next

1)- this element is titanium (atomic number 22)which is d-block element and in 4B group and period 4

2)-this element is nitrogen (atomic number is 7) which is p-block element and in 5A group and 2(second) period.

next

n is known as the principal quantum number

l is known as the azimuthal quantum number

ml is known as the magnetic quantum number

and

n specifies energy and average distance from nucleus(B)

l specifies subshell orbital shape (A)

ml specifies orbital orientation (c)

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if the value of n=2 then l=0 to  l-1

that is if n =2 then l s 0 and 1

total number of orbital can be calculated from 2n that is

22 = 4 orbitals

for l=3 value for ml = -1- 0- +1 that is for l=3 value for ml = -3 to +3

there will be total 7 orbital present at l=3 sublevel because l=3 F orbital(which has 7 sublevels)

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we know that de-broglie wavelength formula is λ = h/mv, value of h(planck constant) = 6.626 * 10-34 m2 Kg/s

  • so for baseball  λ = 6.626 * 10-34/0.140 * 40.2 = 1.177 * 10-34 m (region can not detect wave like properties only particle like properties)
  • for electron λ = 6.626 * 10-34/9.11 * 10-31 * 5.81 * 106 = 0.125 * 10-10 m (region X-Ray)
  • for golf ball λ =6.626 * 10-34 / 4.50 * 10-2 *31.3 = 0.047 * 10-32 m (region can not detect wave like properties only particle like properties)

NEXT

we know that de-broglie wavelength formula is λ = h/mv, value of h(planck constant) = 6.626 * 10-34 m2 Kg/s

  • for bullet λ = 6.6 * 10-34 /1.90 * 10-3 * 313 = 0.011 * 10-31 m (region can not detect wave like properties only particle like properties)​​​​​​​
  • for alpha particle λ = 6.6 * 10-34/ 6.64 *10-27 * 1.52 * 107 =0.656 * 10-14 m (region gamma rays)
  • for electron λ = 6.626 * 10-34/9.11 * 10-31 * 5.81 * 106 = 0.125 * 10-10 m (region X-Ray)
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