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Question: What is an example of a Planck’s constant practice problem?...

What is an example of a Planck’s constant practice problem?

Explanation

Solution

Planck suggested a quantum theory of radiation that marks the atoms and molecules as bundles of energy called as quantum or a photon. The energy of a quantum of radiation is directly proportional to the frequency of radiation. The formula on this basis is E=hνE=h\nu . h is Planck’s constant that has a value 6.626×1034J.s6.626\times {{10}^{-34}}J.s.

Complete answer:
Planck’s quantum theory of radiation explained the particle nature of electromagnetic radiation. According to this theory, atoms and molecules emit energy in a discontinuous pattern in the form of bundles called quantum or photons. The energy of the photons is directly proportional to the frequency of radiation that gave rise to the formula E=hνE=h\nu , where E is the energy in Joules, h is Planck’s constant, and ν\nu is the frequency of radiation calculated in hertz.
Now, example of a Planck’s constant practice problem is:
A red light consists of photons that emit the energy of2.84×1019J2.84\times {{10}^{-19}}J. Calculate their frequency?
As we know, E=hνE=h\nu , where E is given as, h (Planck’s constant) is 6.626×1034J.s6.626\times {{10}^{-34}}J.s, and frequencyν\nu is unknown. So, solving for ν\nu , rearranging the equation we have,
ν=Eh\nu =\dfrac{E}{h}
ν=2.84×1019J6.626×1034J.s\nu =\dfrac{2.84\times {{10}^{-19}}J}{6.626\times {{10}^{-34}}J.s}
ν=4.29×1014s\nu =\dfrac{4.29\times {{10}^{14}}}{s}= 4.29×1014Hz4.29\times {{10}^{14}}Hz.
Hence, the Planck’s practice problem is explained on the basis of E=hνE=h\nu .

Note:
The unit of hertz is the reciprocal of seconds as 1Hz=1s1Hz=\dfrac{1}{s} , therefore the frequency is calculated in hertz as Planck’s constant involves the unit in joule second. Planck's quantum theory became a basis for the photoelectric effect as it involves the ejection of photons. The equation E=hνE=h\nu can be written with the number of photons N as E=NhνE=Nh\nu .