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daser333 [38]
3 years ago
11

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t

he incident particles scatter. The electron in the ground state of a hydrogen atom is found to be momentarily at a distance a0/2 from the nucleus in 1 300 of the observations. In this set of trials, how many times is the atomic electron observed at a distance 2a0 from the nucleus?
Physics
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }

p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }

p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }

N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}

N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } }  }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}

N = 1300(16) e^{-3}

N = 1035.57

N = 1036 times

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<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird  decided to take a pit stop and slows down for 250 m. She spent 5 seconds  in the pit stop.

Here final velocity v=0 \ m/s

initial velocity u= 71 m/s  distance  

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v=u+at

t= \frac {(v-u)}{a}

= \frac {(0-71)}{(-10.082)}=7.042 s

t_1=7.042 s

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After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2

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t= \frac{(71-0)}{6.81}= 10.425 s

t_3=10.425 s

total time= t_1 +t_2+t_3=7.042+5+10.425=22.467 s

Distance covered by the Mercedes Benz during this time is given by s=vt=71 \times 22.467= 1595.157 m

Distance covered by the Thunderbird during this time=250+350=600 m

Difference between distance covered by the Mercedes  and Thunderbird

= 1595.157-600=995.157 m

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

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