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Wittaler [7]
3 years ago
15

Find the energy of one quantum of microwave radiation with frequency 4 GHz and with 500 GHz .

Physics
1 answer:
REY [17]3 years ago
8 0

Answer:

1. 1.65*10^{-5} eV and 2.07*10^{-3}eV 2.2.10*10^{-7}m

Explanation:

The energy of a photon is given by the so-called Planck-Einstein relation:

E=fh=\frac{ch}{\lambda}

where E is the energy, f the frequency, \lambda the wavelength, c the speed of light and h the planck constant

So, we have:

E=fh\\E=(4*10^9Hz)(4.136*10^{-15}eV*s)=1.65*10^{-5} eV\\

And:

E=fh\\E=(5*10^{11}Hz)(4.136*10^{-15}eV*s)=2.07*10^{-3} eV\\

2. The work function is the amount of energy required to remove an electron from the material. Therefore the kinetic energy of the photoelectrons will be the energy of the electromagnetic radiation minus the work function:

K=E-W

Where K is the kinetic energy and W the work function, for copper W=4.7eV

Rewriting for  \lambda:

E=K+W\\\frac{ch}{\lambda}=K+W\\\lambda=\frac{ch}{K+W}\\\lambda=\frac{(3*10^8m/s)4.136*10^{-15}eV}{1.2eV+4.7eV}=2.10*10^{-7}m

This wavelength corresponds to the ultraviolet range

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Not sure the precise concept of "normal observation", but I assume that is observed by "eyes".

Eye observation is basically macroscopic, but when you use a mark, which can be regarded as a point of mass, then it goes to microscopic.

Mark is a reference point which you can compare the relative position change, but with your eyes, first you cannot notice microscopic changes, second the eyes cannot precisely set a stable reference point.
5 0
3 years ago
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A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes th
Alexus [3.1K]

Answer:

a)   v² = v₀² + 2 g h,  b)   t = v₀/g  (1+ √ (1 + 2gh/ v₀²))

Explanation:

a) This is an exercise that we can solve using conservation of energy.

Starting point. High point

         Em₀ = K + U = ½ m v₀² + m gh

Final point. Soil

         Em_{f} = K = ½ m v²

energy is conserved because there is no friction

         Emo = Em_{f}

         ½ m v₀² + m g h = ½ m v²

         v² = v₀² + 2 g h

b) the time it takes to reach the ground can be calculated with kinematics

let's create a reference frame with positive upward direction

         v = vo - g t

when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀

        v = -v₀ - gt

        t = - (v + v₀) / g

we substitute the velocity values ​​calculated in the previous part

        t = - (√(v₀² + 2 g h) + vo) / g

we will simplify the equation a bit

        t = - v₀/g  (1+ √ (1 + 2gh/ v₀²))

c) is now thrown vertically upward with the same initial velocity vo.

   To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a

         v = √ (v₀² + 2gh)

d) for this part if there is change since the speed is not squared

     v₀ = v₀

          v = v₀ - gt

          t = (v₀ - v) / g

          t = (v₀ - √(v₀² + 2 g h)) / g

          t = v₀/g   (1 - √(1 + 2gh / v₀²))

3 0
2 years ago
What is the name of the atomic model in which electrons are treated as waves
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Answer:

The atomic model in which electrons are treated as waves is called the wave mechanical model of the atom or the quantum mechanical model of the atom. Principal energy levels contains energy sublevels

5 0
3 years ago
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e
Gala2k [10]

Explanation:

Given that,

Charge acting on the object, q=-4\ nC=-4\times 10^{-9}\ C

Force acting on the object, F=19\ nC=19\times 10^{-9}\ C (in downward direction)

(a) The electric force acting in the electric field is given by :

F=qE

E is the electric field

E=\dfrac{F}{q}

E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, q=1.6\times 10^{-19}\ C

The force acting on the proton is :

F=qE

F=1.6\times 10^{-19}\times 4.75

F=7.6\times 10^{-19}\ N

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0
3 years ago
A small electronic device is rated at 0.16 W when connected to 115 V. What is the resistance of this device?
Ierofanga [76]

Answer:

150156.25 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of electric current in a circuit. The S.I unit of resistance is Ohm's (Ω)

The expression for resistance is given as

P = V²/R................ equation 1

Where P = power, V = Voltage, R = Resistance.

Making R the subject of the equation,

R = V²/P.................. Equation 2

Given: V = 115 V , P = 0.16 W.

Substitute into equation 2

R = 155²/0.16

R = 150156.25 Ω

Hence,

The resistance = 150156.25 Ω

8 0
2 years ago
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