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seropon [69]
4 years ago
8

In the Bohr model of the hydrogen atom, an electron in the 7th excited state moves at a speed of 4.47 104 m/s in a circular path

having a radius of 2.59 10-9 m. What is the effective current associated with this orbiting electron
Physics
2 answers:
natima [27]4 years ago
8 0

Answer:

4.4 x 10^-7 A

Explanation:

charge on an electron, q = 1.6 x 10^-19 C

velocity, v = 4.47 x 10^4 m/s

radius, r = 2.59 x 10^-9 m

Let the effective current is i.

According to the definition of current, the rate of flow of charge through the conductor is called current.

i = q / T

where, T is the time period of rotation.

T = \frac{2\pi r }{v}

T = \frac{2\times 3.14\times 2.59\times 10^{-9} }{4.47\times 10^{4}}

T = 3.64 x 10^-13 s

So, current

i=\frac{1.6\times 10^{-19}}{3.64\times 10^{-13}}

i = 4.4 x 10^-7 A

tekilochka [14]4 years ago
7 0

Answer:

I = 4.38 x 10⁻⁷ A

Explanation:

Given,

speed = 4.47 x 10⁴ m/s

radius of the circular path, r = 2.59 x 10⁻⁹ m

effective current = ?

The current represented by the orbiting electron is equal to

I = q \times f

f is the frequency

q is charge of electron

we know,

f = \dfrac{v}{2\pi r}

f = \dfrac{4.47\times 10^{4}}{2\pi\times 2.59\times 10^{-9}}

f = 2.74 x 10¹² Hz

now,

I = 1.6 \times 10^{-19}\times 2.74\times 10^{12}

I = 4.38 x 10⁻⁷ A

Hence, the effective current associated with the orbiting electron is equal to I = 4.38 x 10⁻⁷ A

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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

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, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

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d(P.E) =qEd

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Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

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Where,

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So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

\rule[225]{225}{2}

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2 years ago
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