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tino4ka555 [31]
3 years ago
15

The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el

ectron beam impinges on the inside of the picture tube screen.
How many electrons strike the screen each second?
The electrons move with a velocity of 4.0\times10^7\;{\rm m/s}. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 {\rm mm}?
Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam? (Hint: What potential difference produced the field that accelerated electrons? This is an emf.)

Physics
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
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Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

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