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Semmy [17]
3 years ago
9

An electron de-excites from the fourth quantum level to the third and then directly to the first. Two frequencies of light are e

mitted. How do their combined energies compare to the energy of the single frequency that would be emitted by de-excitation from the fourth level directly to the first level?
Physics
1 answer:
4vir4ik [10]3 years ago
6 0

Answer:

The answer is explained below.

Explanation:

The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.

Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.

Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation  from fourth level to ground level directly.

The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.

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saw5 [17]

2) they add the insulation to better the durability or the earbuds

6 0
3 years ago
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To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

5 0
4 years ago
What kind of light results when the total spectrum of refracted light is recombined? violet green white red?
ElenaW [278]
Hello
The final light will be white. In fact, each color of the visible spectrum is an electromagnetic wave with its own specific frequency and wavelength. White, instead, does not have a specific frequency: it is the sum of all the different wavelengths of the visible spectrum. Therefore, when recombining the spectrum of the refracted light all the different frequencies recombine together, and their sum gives white light.

(edited)
4 0
3 years ago
Guys answer with a clear explanation and plzz don't spam.
timama [110]

Answer:

20.7N

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

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3 0
3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
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