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Gala2k [10]
3 years ago
10

The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan

tum number. Imagine a situation where the electron makes a transition from n = 3 to n = 1 state. Where does the photon emitted in this transition lie in the electromagnetic spectrum? (A) visible region (B) Infra-red region (C) Ultra-violet region (D) X-ray region
Physics
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

E_3 =- \frac{13.6 }{3^2}

E_3 =- \frac{13.6 }{9} = -1.511 eV

Energy of the orbit where n = 1 is given as follows

E_1 =- \frac{13.6 }{1^2}

E_1 =- \frac{13.6 }{1} = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

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The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.
bagirrra123 [75]
The radius of the Ferris wheel is r = 83/2 = 91.5 m.

The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s

The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
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From the vector diagram, the change in velocity is (from the Law of Cosines)
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where Ф = Π - 1.4487 = 1.6929 rad
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The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²

The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²

Answer: 
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.

7 0
3 years ago
True or False? A mechanical wave needs a medium to transfer energy. True False
Vlada [557]

Answer:

true I think

Explanation:

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6 0
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The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of
jasenka [17]

1) Frequency: 3.29\cdot 10^{15}Hz

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

E=2.18 aJ=2.18\cdot 10^{-18} J

The energy of a photon is given by

E=hf

where h=6.63\cdot 10^{-34}Js is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz


2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

\lambda=\frac{c}{f}

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5 0
3 years ago
The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.
Vladimir79 [104]
If you do not have to use relative physics but classic physics, this is how you solve it:

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Conversion to km/s = 0.09 * 3.0 * 10^5 km/s = 27000 km/s

Note that this solution, using classic physics do not take into account time and space dilation.

Answer: 27000 km/s
4 0
3 years ago
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