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uranmaximum [27]
4 years ago
8

A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

the mass of a proton. If the HIHI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring?
Physics
1 answer:
vlada-n [284]4 years ago
0 0

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency f=7.0\times10^{13}\ Hz

We need to calculate the reduced mass

Using formula of reduced mass

\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}

Where, m_{H}= atomic mass of H

m_{I}= atomic mass of I

Put the value into the formula

\mu=\dfrac{1\times126.9}{1+126.9}

\mu=0.99\ u

\mu=0.99\times1.66\times10^{-27}\ Kg

\mu=1.643\times10^{-27}\ kg

We need to calculate the force constant of the spring

Using formula of frequency

f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}

k=f^2\times 4\pi^2\times\mu

Put the value into the formula

k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}

k=317.8\ N/m

Hence, The force constant of the spring is 317.8 N/m.

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8 0
4 years ago
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An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other
Umnica [9.8K]
The answer is c. +2.0 µC

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹  N*m/C²


It is given:

F = 7.2 N

d = 0.1 m =  10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

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Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

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Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

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Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

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