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Calculate the reaction force

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The electron moves in an infinitely deep potential well with a width $ℓ = 0.35 nm$.

a) Calculate the electron's minimum speed $v_1$ (i.e. the ground state).

b) Calculate the reaction force, which is causes by electron moving back and forth and colliding to another wall of well adiabatically (thermal insulation).

c) Calculate the frequency of electron's reciprocating motion.

What I did:

a) $E_1=\frac{h^2}{8m_el^2} \hspace{10mm} v_1=\sqrt{\frac{2E_1}{m_e}}$

So I got this answer, which is correct:

$E_1=4.9176...\times10^{-19}\ J$

$v_1=1.0390...\times10^6\ m/s\approx1.0\times10^6\ m/s$

b) I tried this formula:

$F_r=\frac{m_ev^2}{r}\hspace{10mm}r_1=\frac{h^2\epsilon_0}{\pi m_ee^2}$

But I didn't get correct answer.

c) $f=\frac{2E_1}{h}$

So I got this answer, which is correct:

$f=1.4843\times 10^{15}\frac{1}{s}\approx 1.5\times 10^{15}\frac{1}{s}$

**So, my question is what formula should I use to get the correct force ($F$ ) value?
Because obviously the formula which I chose is incorrect.

EDIT:

I used this formula and got correct answer:

$F=\frac{\triangle p}{\triangle t}=\frac{2m_ev_1}{2\frac{l}{v_1}}=m_ev_1\times\frac{v_1}{l}=\frac{m_ev_1^2}{l}\approx 2.8\times 10^{-9}\ N$

So is it correct approach? Or I was just lucky and got correct answer?

asked Mar 17, 2017 in Physics Problems by RedRose (140 points)
edited Mar 18, 2017 by RedRose
Your method is correct.

1 Answer

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Best answer

The formulas you are using for force and radius (?) are for the Bohr model of the atom. That is a 2D model for circular motion (continuous acceleration). The question concerns a 1D model of a particle bouncing between the walls of a potential well. There is only acceleration when the particle hits the walls.

The particle rebounds elastically from the walls, so the change in momentum is $2mv$ for each collision. Force is rate of change of momentum so you need to multiply this by the frequency of collisions with one of the 2 walls.

answered Mar 18, 2017 by sammy gerbil (26,482 points)
selected Mar 18, 2017 by RedRose
Thank you :) But see my "edit" in my original Q.
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