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3 years ago

PLEASE HELP ME I AM TIMED!

Physics

2 answers:

enyata [817]3 years ago

8 0

I believe The answer is B

anyanavicka [17]3 years ago

3 0

Is has to be A or B but I think it’s A

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3 years ago

A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

postnew [5]

Answer:

r=15.53 nm

$F=9.57\times 10^{-13}N$

Explanation:

Lets take electron is in between iron and uranium

Charge on electron $q_1= -1.602\times 10^{-19}C$

Charge on iron $q_2= 2\times 1.602\times 10^{-19}C$

Charge on uranium $q_3= 1.602\times 10^{-19}C$

We know that force between two charge

$F=K\dfrac{q_1 q_2}{r^2}$

$K=9\times 10^9\dfrac{N-m^2}{c^2}$

For equilibrium force between electron and iron should be force between electron and uranium

Lets take distance between electron and uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

$K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}$

$K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}$

$q_2= 2\timesq_1,q_3=q_1$

$\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}$

So r=15.53 nm

So force

$F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}$

$F=9.57\times 10^{-13}N$

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