According to Max Planck energy of photon is

According to Planck's quantum theory of light, every source of radiation emits photons (i.e., packets of energy ) which travel in all directions with the same speed (i.e., speed of light). Each photon is of energy E=hv=hc//lambda , where h is Planck's constant, v is the frequency and lambda is wavelength of radiation emitted. The photons emitted from different sources of radiation are different. Read the above passage and answer the following questions: (i) on what factors does the number of photons emitted per second from a source of radiations depends? (ii) Why are the high energy photons not vissible to eye? (iii) Which basic values do you learn from this study?

According to the Planck, the radiation is emitted by a body discontinuously in the form of small packets called

The de-Broglie wavelength of an electron is same as the wavelength of a photon. The energy of a photon is ‘x’ times the K.E. of the electron, then ‘x’ is: (m-mass of electron, h-Planck’s constant, c - velocity of light)

The de-Broglie wavelength of an electron is same as the wavelength of a photon. The energy of a photon is ‘x’ times the K.E. of the electron, then ‘x’ is: (m-mass of electron, h-Planck’s constant, c - velocity of light)

Write equation of energy of photon.

According to planck's quantum theory any electromagnetic radiation is