a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin
1 answer:
This question involves the concepts of Wein's displacement law and characteristic wavelength.
The blackbody temperature will be "3.22 x 10⁵ k".
<h3>WEIN'S DISPLACEMENT LAW</h3>According to Wein's displacement law,
where,
= characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
- T = temperature = ?
- c = Wein's displacment constant = 2.897 x 10⁻³ m.k
Therefore,
T = 3.22 x 10⁵ k
Learn more about characteristic wavelength here:
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We can solve the problem by using the law of conservation of energy.
Initially, the rock has only gravitational potential energy, which is given by
where mg is the weigth of the rock (2200 N), while h is the height at which the rock has been released (h=15 m). If we calculate it, we get
Just before hitting the ground, the rock height is zero, so its potential energy is now zero. So the total mechanical energy of the rock now is just kinetic energy:
however, the mechanical energy of the rock must be conserved, so
and so we have that the kinetic energy of the rock just before hitting the ground is equal to its initial potential energy:
Initially, the rock has only gravitational potential energy, which is given by
where mg is the weigth of the rock (2200 N), while h is the height at which the rock has been released (h=15 m). If we calculate it, we get
Just before hitting the ground, the rock height is zero, so its potential energy is now zero. So the total mechanical energy of the rock now is just kinetic energy:
however, the mechanical energy of the rock must be conserved, so
and so we have that the kinetic energy of the rock just before hitting the ground is equal to its initial potential energy: