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UNO [17]
3 years ago
5

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro

ximately 2.15 106 m/s?
Chemistry
1 answer:
Zolol [24]3 years ago
8 0

This is an incomplete question, here is a complete question.

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.15\times 10^6m/s?

Find the force acting on the electron as it revolves in a circular orbit of radius 5.34\times 10^{-11}m.

Answer : The force acting on electron is, 7.9\times 10^{-8}N

Explanation :

Formula used :

F=\frac{mv^2}{r}

where,

F = force acting on electrons

m = mass of electrons = 9.1\times 10^{-31}kg

v = speed of electron = 2.15\times 10^6m/s

r = radius of circular orbit = 5.34\times 10^{-11}m

Now put all the given values in the above formula, we get:

F=\frac{(9.1\times 10^{-31}kg)\times (2.15\times 10^6m/s)^2}{5.34\times 10^{-11}m}

F=7.9\times 10^{-8}N

Thus, the force acting on electron is, 7.9\times 10^{-8}N

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IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = \frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M

Concentration of Br_{2} = \frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M

Concentration of NOBr = \frac{0.102}{1}M=0.102M

Reaction quotient,Q_{c} , for this reaction = \frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}

species inside third bracket represents concentrations at the given interval.

So, Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106

So, the reaction is not at equilibrium.

As Q_{c}< K_{c} therefore reaction must run in forward direction to increase Q_{c} and make it equal to K_{c}.

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