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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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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c
stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

8 0
3 years ago
A rock falls from the top of a hill to the bottom. The total energy of the rock at the top isn’t the same as the sum of its pote
mart [117]

Answer:

A) some of the rocks energy is transformed to thermal energy

Explanation:

If we neglect air resistance during the fall of the rock, than the mechanical energy of the rock (which is sum of its potential energy and its kinetic energy) would be constant during the entire motion, so the total energy of the rock at the top would be the same as the sum of its potential energy and kinetic energy at the bottom.

However, this not occurs, due to the presence of air resistance. In fact, air resistance acts against the fall of the rock, and because of the friction between the molecules of air and the surface of the rock, the rock loses part of its energy. This energy is converted into thermal energy of the molecules of the air.

3 0
4 years ago
Within a vacuum, the property to all electromagnetic waves is their
bezimeni [28]
The answer is D) Velocity
7 0
4 years ago
Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o
Rom4ik [11]
Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
178/4.905=m

His mass is 36.3 kg
4 0
3 years ago
Read 2 more answers
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom
jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

3 0
3 years ago
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