Question

Wavelength of certain radiation is 6625 angstrom then the number of quantas present in 1joule of energy of that radiation is

Answer 1

Answer:

The wavelength of monochromatic radiation =6625A˚

Energy of photon, E=λhc​

                              =6.623×10−106.626×10−34×3×108​

                              =3×10−19Joule

Photon emitted per second=3×101960​

                                             =2×1020  

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Answer 2

Answer:

The number of quanta present in 1joule of energy of that radiation is=> 3.332×10^18 Photons.

Explanation:

According to Planck's theory, emission of radiation is discontinuous in discrete packets called quanta.

From Planck’s Equation energy emitted by radiation is quantized E=Nhv

where, N=1,2,3,4...........

Energy of 1 photon = hc/λ

λ=wavelength=6625А°=6625×10∧-10m

C= velocity of light in vacuum=3×10∧8m/s

∴ the energy of photon(E)=(6.626×10∧-34×3×10∧8)/(6.625×10∧-10)

                                          = 3×10∧-19J

Now 3×10∧-19J is the energy 1 photon

∴1J will be the energy of 1/3×10∧-19=3.332×10∧18 photons.

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