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3 years ago

How are atomic emission spectra like fingerprints for the elements

1 answer:

8 0

Atomic emission spectra are like fingerprints for the elements, because it can show the number of orbits in that elements as well as the energy levels of that element. As each emission of atomic spectra is unique, it is the fingerprint of element.

<u>Explanation: </u>

Each element has unique arrangement of electrons in different energy levels or orbits. So depending upon the difference in energy of the orbital, the emission spectra will be varying for each element. As the binding energy and excitation energy is not common for any two elements, so the spectra obtained when those excited electrons will release energy to ground state will also be unique.

As in atomic emission spectra, the incident light will be absorbed by the electrons of those elements making the electron to excite, then the excited electron will return to ground state on emission of radiation of energy. Thus, this energy of emission is equal to the difference between the energy of initial and final orbital. So the spectra will act like fingerprints for elements.

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A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the osc

ludmilkaskok [199]

Answer:

.487 s⁻¹

Explanation:

Let damping constant be τ . The equation of decreasing amplitude can be written as

A = A₀ $e^{-\tau t$

A / A₀ = $e^{-\tau t$

At t = 9.43 s , A / A₀ = .01

.01 = $[e^{-\tau\times9.43$

ln.01 = - 9.43 τ

-4.6 = -9.43τ

τ = .487 s⁻¹

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4 years ago

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trapecia [35]

Answer:

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Explanation:

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3 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

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Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

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3 years ago

A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is 5) its radius? (The value of o is 4 ×

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The self-inductance of a solenoid is given by:
$L= \frac{\mu_0 N^2 A}{l}$
where
$\mu_0$ is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid

For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
$A= \frac{Ll}{\mu N^2}= \frac{(0.025 H)(0.70 m)}{(4\pi \cdot 10^{-7}N/A^2)(3000)^2}= 1.55 \cdot 10^{-3}m^2$
And since the area is related to the radius by
$A=\pi r^2$
The radius of the solenoid is
$r= \sqrt{ \frac{A}{\pi} } = \sqrt{ \frac{1.55 \cdot 10^{-3} m^2}{\pi} } =0.022 m=2.2 cm$

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3 years ago

6. Name two ways in which Faraday's work affects your own life in the twenty-first century.<br><br>

Gekata [30.6K]

He discovered electromagnetic induction

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4 years ago