Question

vYou may have done these temperature calculations before, but it is useful to practice with these concepts, and this time you'll use the temperatures to determine the heat capacity.) The interatomic spring stiffness for gold is determined from Young's modulus measurements to be 20 N/m. The mass of one mole of gold is 0.197 kg. If we model a block of gold as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)

Answer 1

Answer:

Answer a=3.423582303 X 10-21 J

Answer C = 5.3036 X 10-23 J/K/atom

Explanation:

You may have done these temperature calculations before, but it is useful to practice with these concepts, and this time you'll use the temperatures to determine the heat capacity.) The interatomic spring stiffness for gold is determined from Young's modulus measurements to be 20 N/m. The mass of one mole of gold is 0.197 kg. If we model a block of gold as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)

Answer 2

The general value of one quantum of energy for any of the oscillators =

3.4236 * 10⁻²¹ J

Given data :

Avogadro's number =  6.0221 * 10²³

Determine the value of one quantum of energy for one of these oscillators

Applying the formula below

$E_{quantum}$ ( hw )  = $h \sqrt{\frac{4kNa}{M} }$  ------- ( 1 )

where : h = 1.0546 * 10⁻³⁴ J.s ,  k = 28 ,  Na = 6.0221 * 10²³ , M = 0.064

Insert values into equation ( 1 )

$E_{quantum}$  ≈ 3.4236 * 10⁻²¹ J ( general value )

But The value of one quantum of energy for each of the oscillators

$E_{20} =$  20 * 3.4236 * 10⁻²¹  ≈ 6.8472 * 10⁻²⁰ J

E₂₁  = 21 * 3.4236 * 10⁻²¹  ≈ 7.190 * 10⁻²⁰ J

E₂₂ = 22 * 3.4236 * 10⁻²¹  ≈ 7.532 * 10⁻²⁰ J

Hence we can conclude that The general value of one quantum of energy for any of the oscillators = 3.4236 * 10⁻²¹ J

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