Question 6: Find energy of each of the photons which
(i) correspond to light of frequency 3× 1015 Hz.
(ii) have wavelength of 0.50 Å.
Answer:
Planck’s quantum theory = Based on the assumption that all atoms on the surface of the heated solid vibrate at the frequency, Planck developed a model that came to be known as Planck’s equation. Through experiments of frequencies and temperature, Planck was able to generate a constant, Planck’s constant
h = 6.62607 x 10-34 J s
Using this constant he was able to restate his theory: energy was directly proportional to frequency. He wrote his equation as
E=hν
where E is energy, h is Planck’s constant, and v is frequency.
(i) Energy (E) of a photon is given by the expression,
E=hν
Where, h = Planck’s constant = 6.626 × 10–34 Js
ν = frequency of light = 3 × 1015 Hz
Substituting the values in the given expression E = hv we get
E = (6.626 × 10–34) (3 × 1015)
E = 1.988 × 10–18 J
(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,
where , h = Planck’s constant = 6.626 × 10–34 Js
c = velocity of light in vacuum = 3 × 108 m/s
Substituting the values in the given expression of E:

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You explained very simply...... so this is very usefull.... thank you alðð
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