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 2.03 106 m/s.(a) Find the force acting on the electron as it revolves in a circular orbit of radius 5.35 10-11 m.
whats the magnitude in N ?

(b) Find the centripetal acceleration of the electron.
whats the magnitude in m/s2

Answers

Answer 1

Answer: In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.

Ok, Rameses:

Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton

Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg

a). The force between two charges is

F = (9 x 10⁹) Q₁ Q₂ / R²

= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

      = ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

=    8.05 x 10⁻⁸ Newton .

b). Centripetal acceleration =

   v² / r .

A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

   =    7.7 x 10²² m/s² .

That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .

I'm going to leave that step for you.   Good luck !

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A spring gains 2.34 joules of elastic potentialenergy as it is compressed 0.250 meter from its
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Answers

The correct answer to the question is: 4) 74.9 N/m.

EXPLANATION:

As per the question, the stretched length of the spring is given as x = 0.250 m.

The potential energy gained by the spring is given as 2.34 joules.

We are asked to calculate the spring constant of the spring.

The potential energy gained by the spring is nothing else than the elastic potential energy .

The elastic potential energy of the spring is calculated as -

Potential energy P.E = $(1)/(2)kx^2$

⇒k = $(2P.E)/(x^2)$

= $(2* 2.34)/((0.250)^2)\ N/m$

= 74.88 N/m

= 74.9 N/m. [ans]

Hence, the force constant of the spring is 74.9 N/m.

Elastic energy equals 1/2 k(spring constant) *Δx^2(change in position) so you set the energy of the spring equal to the 1/2kΔx^2 and solve for K. 2.34=1/2k(.250)^2 which gives you a spring constant of 74.9 N/m.

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Can u please solve and then explain answer