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fgiga [73]
3 years ago
14

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

magnetic field isperpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 700V.
Part A:What is the electron's speed?(Express your answer to two significant figures and include the appropriate units.)

Part B: If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field?(Express your answer to two significant figures and include the appropriate units.)
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

v=15689290.81\ m/s

v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

r=0.042\ m

r=4.2\ cm

Hence, (a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2 cm

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Answer:

1.87 A

Explanation:

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A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

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4 0
3 years ago
Fluorescence occurs when an object absorbs ____________________ and immediately releases the energy as visible light.
soldier1979 [14.2K]

Answer:

The answer should be light or other electromagnetic radiation

Explanation:

Such as x-rays or other things like that.

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Answer:

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Explanation:

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\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y

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Answer:

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