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Gnesinka [82]
2 years ago
5

Physicists often use a different unit of energy, the electron volt, when dealing with energies at the atomic level. One electron

volt, abbreviated eV, is defined as the amount of kinetic energy gained by an electron upon accelerating through a 1.0 V potential difference.
a) what is 1.0 electron volt in joules?

b) what is the speed of a proton with 5000 eV of kinetic energy?
Physics
1 answer:
Alex Ar [27]2 years ago
3 0

Answer:

a)  E = 1.06 10⁻¹⁹ J,     b)   v = 9.78 10⁵ m / s

Explanation:

The physical magnitudes can be given in several units, but in general all must be reduced to the same system in a given exercise, the most used system is the international (SI)

           1 eV =q V=  1.6 10⁻¹⁹ J

let's reduce the quantities requested

a) E = 1.0 eV to Joule

           E = 1.0 eV (1.6 10⁻¹⁹ J / 1 eV)

           E = 1.06 10⁻¹⁹ J

b) the kinetic energy is given by

           K = ½ m v²

           v = \sqrt{\frac{2K}{m} }

the mass of the proton is

          m = 1,673 10⁻²⁷ kg

let's reduce the energy to the SI system

           E = 5000 ev (1.6 10-19 J / 1 eV) = 8000 10⁻¹⁹ J

let's calculate

           v = \sqrt{ \frac{2 \ 8000 \ 10^{-19}}{1.673 \ 10^{-27} }  }

           v = \sqrt{ 95.637 \  10^{10} }

           v = 9.78 10⁵ m / s

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Answer:

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Explanation:

The speed of a wave can be defined as:

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Where v is the speed, \nu is the frequency and \lambda is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

c = \nu \cdot \lambda (2)              

 

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Therefore, \lamba\lambda can be isolated from equation 2 to get the wavelength of the photon.

\lambda = \frac{c}{\nu} (3)

\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}

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Hence, the photon has a wavelength of 1.5x10^{-6}m        

<em>Summary:  </em>

Photons are the particles that constitutes light.

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If the half life of an isotope is 20 years, how much of the original amount will
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6 0
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A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

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An electron experiences a downward force of 12.8×10-19 N while traveling in a magnetic field of 8×10-5 T west, what is the magni
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Answer:

v=10^5\ m/s

Explanation:

Given that,

Magnetic force acting on an electron, F=12.8\times 10^{-19}\ N

The magnitude of the magnetic field,B=8\times 10^{-5}\ T

We need to find the magnitude of the velocity. We know that the magnetic force is given by :

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v=\dfrac{F}{qB}\\\\v=\dfrac{12.8\times 10^{-19}}{1.6\times 10^{-19}\times 8\times 10^{-5}}\\\\v=10^5\ m/s

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