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Pani-rosa [81]
3 years ago
7

In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of

the electron when it is in the ground state. rev/s (b) Determine the orbital frequency of the electron when it is in the first excited state.
Physics
1 answer:
Airida [17]3 years ago
3 0

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

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Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

               (55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>


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An air mass is a large pocket of air with _____.
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Consider an electric dipole in a uniform electric field. In which orientation does the dipole-field system have the greatest pot
Gekata [30.6K]

Answer: when the dipole moment and electric field are parallel

Explanation: The formulae that relates the potential energy of a dipole, dipole moment and strength of electric field is given as

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Write about Archimedes principle​
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8 0
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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
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