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Question: If the interatomic spacing in a steel wire is 3.0Å and \(Y _ { \text {steel } }\)= <img src="https:...

If the interatomic spacing in a steel wire is 3.0Å and

Ysteel Y _ { \text {steel } }= then force constant is

A

6×10−2N/A˚6 \times 10 ^ { - 2 } N / \AA

B

6×10−9N/A˚6 \times 10 ^ { - 9 } N / \AA

C
D
Answer

6×10−9N/A˚6 \times 10 ^ { - 9 } N / \AA

Explanation

Solution

K=Yr0=20×1010×3×10−10=60 N/mK = Y r _ { 0 } = 20 \times 10 ^ { 10 } \times 3 \times 10 ^ { - 10 } = 60 \mathrm {~N} / \mathrm { m } =6×10−9N/A˚= 6 \times 10 ^ { - 9 } N / \AA