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3 years ago

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Fe has the electron configuration [Ar]3d64s2. What is the magnitude of the orbital angular momentum for its most energetic elect

ron? √2h Võh - V12h V12h 20h

1 answer:

salantis [7]3 years ago

7 0

Answer : The magnitude of the orbital angular momentum for its most energetic electron is, $\sqrt{6}\hbar$

Explanation :

The formula used for orbital angular momentum is:

$L=\sqrt{l(l+1)}\hbar$

where,

L = orbital angular momentum

l = Azimuthal quantum number

As we are given the electronic configuration of Fe is, $[Ar]3d^64s^2$

Its most energetic electron will be for 3d electrons.

The value of azimuthal quantum number(l) of d orbital is, 2

That means, l = 2

Now put all the given values in the above formula, we get:

$L=\sqrt{2(2+1)}\hbar$

$L=\sqrt{6}\hbar$

Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, $\sqrt{6}\hbar$

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Answer:

non linear square relationship

Explanation:

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a = mv^2/r

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3 years ago

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati

Afina-wow [57]

Answer:

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Explanation:

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Putting the values

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option d is correct.

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3 years ago

Describing a Physical Change

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Answer:

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2 years ago

A baseball of mass m = 0.145 kg is suspended vertically from a tree by a string of length L = 1.1 m and negligible mass. Take z

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3 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

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3 years ago