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zlopas [31]
3 years ago
8

A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electro

des in the tube. The gas ionizes; electrons move toward the positive terminal and singly charged positive ions move toward the negative terminal. What is the current in a hydrogen discharge tube in which 3.4 ✕ 1018 electrons and 1.4 ✕ 1018 protons move past a cross-sectional area of the tube each second? (Enter the magnitude.)
Physics
1 answer:
pishuonlain [190]3 years ago
8 0

Explanation:

It is given that the number of electrons passing through the cross-sectional area in 1 s is 3.4 \times 10^{18}. Also, we know that charge on an electron is -1.60 \times 10^{-19} C, then negative charge crossing to the left per second is  as follows.

         I- = 3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons

         I- = 0.544 A

As it is given that the number of protons crossing per second is 1.4 \times 10^{18}, as the charge on the proton is +1.60 \times 10^{-19} C, then positive charge crossing to the right per second is calculated as follows.

          I+ = 1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C

            I+ = 0.224 A

          I = l I+ l + l I- l

So,    I = 0.544 + 0.224

            = 0.768 A

Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

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D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
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