JEE MAIN - Physics (2018 (Offline) - No. 11)
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)
5.5 N/m
6.4 N/m
7.1 N/m
2.2 N/m
Explanation
6.02 $$ \times $$ 1023 atoms of silver = 108 gm
1 atoms of silver = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ kg
For a harmonic oscillator
f = $${1 \over {2\pi }}$$ $$\sqrt {{k \over m}} $$
Where k = force constant
$$ \Rightarrow $$$$\,\,\,$$ f2 = $${1 \over {4{\pi ^2}}}$$ $$\left( {{k \over m}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ k = mf2 $$ \times $$ 4$$\pi $$2
Given,
f = 1012
m = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$
$$\therefore\,\,\,$$ k = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ $$ \times $$ 1012 $$ \times $$ 4$$\pi $$2
= 7.1 N/m
1 atoms of silver = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ kg
For a harmonic oscillator
f = $${1 \over {2\pi }}$$ $$\sqrt {{k \over m}} $$
Where k = force constant
$$ \Rightarrow $$$$\,\,\,$$ f2 = $${1 \over {4{\pi ^2}}}$$ $$\left( {{k \over m}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ k = mf2 $$ \times $$ 4$$\pi $$2
Given,
f = 1012
m = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$
$$\therefore\,\,\,$$ k = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ $$ \times $$ 1012 $$ \times $$ 4$$\pi $$2
= 7.1 N/m
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