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DaniilM [7]
3 years ago
8

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

The ion returns to the ground state by emitting FOUR photons ONLY. What is the wavelength of the second highest energy photon?
Physics
1 answer:
Xelga [282]3 years ago
6 0

Answer:

\lambda=1282\ nm

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=5\ and\ n_f=3

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m

\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m

\lambda= 1.2828\times10^{-6}

1 m = 10⁻⁹ nm

\lambda=1282\ nm

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1/f = (l + p) / (pl)

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Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre
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Answer:

d= 0.242 mm

Explanation:

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Here n = 2

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The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

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m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

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I = \frac{15}{32}\cdot M\cdot R^{2}

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Please see this question related to Moments of Inertia: brainly.com/question/15246709

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