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Vera_Pavlovna [14]
2 years ago
14

An 11.8 eV photon is absorbed by an electron in an atom, causing the electron to be excited from level 1 to level 3. The electro

n then emits a 4.1 eV photon and shifts to level 2. What photon will be emitted when the electron moves back to level 1?
a. 15.9 eV
b. 11.8 eV
c. 7.7 eV
d. 4.1 eV
Physics
1 answer:
densk [106]2 years ago
6 0

Answer:

c. 7.7 eV

Explanation:

The computation of the photon emitted is shown below:

Given that

From level 1 to level 3, the energy absorbed by electrons is 11.8eV

And, when from level 3 to level3, the energy released by electron is 4.1eV

So, from level 2 to level 1, the energy released is

= 11.8 - 4.1

= 7.7

= 7.7eV

Hence, the third option is correct

The same is to be considered

And, the other options are wrong

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Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
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El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

Vf ^{2} -Vo ^{2} =2 x g x h

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2}  \\  \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2}  }  \\ Vf=28,64 m/s

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How large is the area of water found on Mars?
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Three cars with identical engines and tires start from rest, and accelerate at their maximum rate. Car X is the most massive, an
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Explanation:

Given

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3 years ago
A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
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