JEE Main On 08 April 2018 Question 21
Question: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $ 10^{12}/\sec $ . What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number $ \text{=6}\text{.02}\times 1{0^{23}}\text{ gm mol}{e^{\text{-1}}} $ ). [JEE Main Online 08-04-2018]
Options:
A) $ 2.2N/m $
B) $ 5.5N/m $
C) $ 6.4N/m $
D) $ 7.1N/m $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ kx=m{{\omega }^{2}}X $
$ \Rightarrow k=m{{\omega }^{2}} $ $ =\frac{108}{6.02\times 10^{23}}\times {{(2\pi \times 10^{12})}^{2}}\times {10^{-3}} $ $ k\approx 7.1N/m $
