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If an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple
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Quoting from the Feynman Lectures on Physics - Vol I:
The atoms are 1 or $2 times 10^−8 rm cm$ in radius. Now $10^−8 rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple.
How does this hold true?
Let us assume the radius of an average apple is about $6 rm cm$ ($0.06 rm m$). The radius of earth is about $6371 rm km$ ($6371000 rm m$). Therefore, a $frac6371000 mathrmm0.06 mathrmm = 106183333.33$ magnification, i.e., a magnification of about $10^-8$ times is required to magnify an apple to the size of the earth.
If we magnify an atom of size say 1 angstrom ($10^-10 rm m$) by $106183333.33$ times, we get $0.0106 rm m$ or $1.06 rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good?
atomic-physics atoms estimation textbook-erratum order-of-magnitude
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Quoting from the Feynman Lectures on Physics - Vol I:
The atoms are 1 or $2 times 10^−8 rm cm$ in radius. Now $10^−8 rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple.
How does this hold true?
Let us assume the radius of an average apple is about $6 rm cm$ ($0.06 rm m$). The radius of earth is about $6371 rm km$ ($6371000 rm m$). Therefore, a $frac6371000 mathrmm0.06 mathrmm = 106183333.33$ magnification, i.e., a magnification of about $10^-8$ times is required to magnify an apple to the size of the earth.
If we magnify an atom of size say 1 angstrom ($10^-10 rm m$) by $106183333.33$ times, we get $0.0106 rm m$ or $1.06 rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good?
atomic-physics atoms estimation textbook-erratum order-of-magnitude
edited 1 hour ago
Chair
3,06341732
asked 1 hour ago
Lone Learner
1111
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1
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago
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up vote
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Quoting from the Feynman Lectures on Physics - Vol I:
The atoms are 1 or $2 times 10^−8 rm cm$ in radius. Now $10^−8 rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple.
How does this hold true?
Let us assume the radius of an average apple is about $6 rm cm$ ($0.06 rm m$). The radius of earth is about $6371 rm km$ ($6371000 rm m$). Therefore, a $frac6371000 mathrmm0.06 mathrmm = 106183333.33$ magnification, i.e., a magnification of about $10^-8$ times is required to magnify an apple to the size of the earth.
If we magnify an atom of size say 1 angstrom ($10^-10 rm m$) by $106183333.33$ times, we get $0.0106 rm m$ or $1.06 rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good?
atomic-physics atoms estimation textbook-erratum order-of-magnitude
edited 1 hour ago
Chair
3,06341732
asked 1 hour ago
Lone Learner
1111
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Lone Learner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Quoting from the Feynman Lectures on Physics - Vol I:
The atoms are 1 or $2 times 10^−8 rm cm$ in radius. Now $10^−8 rm cm$ is called an angstrom (just as another name), so we say they are 1 or 2 angstroms (Å) in radius. Another way to remember their size is this: if an apple is magnified to the size of the earth, then the atoms in the apple are approximately the size of the original apple.
How does this hold true?
Let us assume the radius of an average apple is about $6 rm cm$ ($0.06 rm m$). The radius of earth is about $6371 rm km$ ($6371000 rm m$). Therefore, a $frac6371000 mathrmm0.06 mathrmm = 106183333.33$ magnification, i.e., a magnification of about $10^-8$ times is required to magnify an apple to the size of the earth.
If we magnify an atom of size say 1 angstrom ($10^-10 rm m$) by $106183333.33$ times, we get $0.0106 rm m$ or $1.06 rm cm$ only. The atom has not been magnified to the size of the original apple. How does the quoted statement in the book hold good?
atomic-physics atoms estimation textbook-erratum order-of-magnitude
atomic-physics atoms estimation textbook-erratum order-of-magnitude
edited 1 hour ago
Chair
3,06341732
asked 1 hour ago
Lone Learner
1111
New contributor
Lone Learner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Chair
3,06341732
asked 1 hour ago
Lone Learner
1111
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Lone Learner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Chair
3,06341732
edited 1 hour ago
Chair
3,06341732
edited 1 hour ago
Chair
3,06341732
3,06341732
asked 1 hour ago
Lone Learner
1111
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asked 1 hour ago
Lone Learner
1111
asked 1 hour ago
Lone Learner
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1111
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1
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago
add a comment |Â
1
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago
1
1
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago
add a comment |Â
2 Answers
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oldest
votes
up vote
3
down vote
When dealing with such orders of magnitude, a factor of 6 can be easily disregarded. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.
answered 50 mins ago
lr1985
36619
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
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up vote
1
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In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. No all apples have the same size, and no all atoms have the same "size". All we can then work with is orders of magnitude, so 1 vs. 6 would be considered to be the "same", since we could be dealing with apples and atoms of certain sizes.
Indeed, if an apple's radius is on the order of $10^-2 rm m$, and if the Earth's radius is on the order of $10^6 rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.
If our atom radius is on the order of one Angstrom, or $10^-10 rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^-2 rm m$, which is what we were trying to show in the first place.
You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the actual size of an atom is on the order of $10^-10 rm m$.
Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so I would say the book is correct.
answered 20 mins ago
Aaron Stevens
3,3391422
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
When dealing with such orders of magnitude, a factor of 6 can be easily disregarded. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.
answered 50 mins ago
lr1985
36619
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
add a comment |Â
up vote
3
down vote
When dealing with such orders of magnitude, a factor of 6 can be easily disregarded. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.
answered 50 mins ago
lr1985
36619
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
When dealing with such orders of magnitude, a factor of 6 can be easily disregarded. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.
answered 50 mins ago
lr1985
36619
When dealing with such orders of magnitude, a factor of 6 can be easily disregarded. Furthermore, note that apples have radii that are often smaller than $6$ cm (see for instance here). If we take an apple with a radius of $3$ cm and an atom with a radius of $2$ angstrom, the "Feynman comparison" (as per your numbers) comes out to be fairly good.
answered 50 mins ago
lr1985
36619
answered 50 mins ago
lr1985
36619
answered 50 mins ago
lr1985
36619
answered 50 mins ago
lr1985
36619
36619
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
add a comment |Â
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
1
1
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
Heh. Feynman, originally from Far Rockaway, may be more accustomed to apples that grow in colder northern regions than those found in warmer southern regions. Thus, his estimate of typical apple size may be smaller. <big wide grin>
– puppetsock
22 mins ago
1
1
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
@puppetsock I wonder how Feyman's apples compared to Newton's apples...
– Aaron Stevens
18 mins ago
add a comment |Â
up vote
1
down vote
In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. No all apples have the same size, and no all atoms have the same "size". All we can then work with is orders of magnitude, so 1 vs. 6 would be considered to be the "same", since we could be dealing with apples and atoms of certain sizes.
Indeed, if an apple's radius is on the order of $10^-2 rm m$, and if the Earth's radius is on the order of $10^6 rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.
If our atom radius is on the order of one Angstrom, or $10^-10 rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^-2 rm m$, which is what we were trying to show in the first place.
You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the actual size of an atom is on the order of $10^-10 rm m$.
Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so I would say the book is correct.
answered 20 mins ago
Aaron Stevens
3,3391422
add a comment |Â
up vote
1
down vote
In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. No all apples have the same size, and no all atoms have the same "size". All we can then work with is orders of magnitude, so 1 vs. 6 would be considered to be the "same", since we could be dealing with apples and atoms of certain sizes.
Indeed, if an apple's radius is on the order of $10^-2 rm m$, and if the Earth's radius is on the order of $10^6 rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.
If our atom radius is on the order of one Angstrom, or $10^-10 rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^-2 rm m$, which is what we were trying to show in the first place.
You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the actual size of an atom is on the order of $10^-10 rm m$.
Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so I would say the book is correct.
answered 20 mins ago
Aaron Stevens
3,3391422
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. No all apples have the same size, and no all atoms have the same "size". All we can then work with is orders of magnitude, so 1 vs. 6 would be considered to be the "same", since we could be dealing with apples and atoms of certain sizes.
Indeed, if an apple's radius is on the order of $10^-2 rm m$, and if the Earth's radius is on the order of $10^6 rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.
If our atom radius is on the order of one Angstrom, or $10^-10 rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^-2 rm m$, which is what we were trying to show in the first place.
You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the actual size of an atom is on the order of $10^-10 rm m$.
Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so I would say the book is correct.
answered 20 mins ago
Aaron Stevens
3,3391422
In "back of the envelope" calculations like this, all you can really do is look at orders of magnitude. No all apples have the same size, and no all atoms have the same "size". All we can then work with is orders of magnitude, so 1 vs. 6 would be considered to be the "same", since we could be dealing with apples and atoms of certain sizes.
Indeed, if an apple's radius is on the order of $10^-2 rm m$, and if the Earth's radius is on the order of $10^6 rm m$, then our "conversion factor" (or as you say magnification factor) is on the order of $10^8$.
If our atom radius is on the order of one Angstrom, or $10^-10 rm m$, then applying our conversion factor when we blow up our apple, we get that the atom is on the order of $10^-2 rm m$, which is what we were trying to show in the first place.
You have to keep in mind that this is quoted as "a way to remember their size". Therefore, if we did not know the size of an atom, we could take the size of an apple, the size of the earth, and then use the derived $10^8$ factor to find that the actual size of an atom is on the order of $10^-10 rm m$.
Feynman is not saying "Take any apple and blow it up to the size of the earth. You will find that the larger atoms are exactly the same size as the original apple". This is purely a memory tool, or it is also a way to describe the size of an atom using more familiar objects. Also the book even uses the word "approximately", so I would say the book is correct.
answered 20 mins ago
Aaron Stevens
3,3391422
answered 20 mins ago
Aaron Stevens
3,3391422
answered 20 mins ago
Aaron Stevens
3,3391422
answered 20 mins ago
Aaron Stevens
3,3391422
3,3391422
add a comment |Â
add a comment |Â
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1
I would say that is is still a quite good approximation. It would be a great way to visualize it (for me) if I had any real sense of the size of the Earth.
– user190081
1 hour ago