If a source of power `4 kW` produces `10^(20)` photons`//`second, the radiation belongs to a part of the spectrum called:

Options

A. X-rays
B. ultraviolet rays
C. microwaves
D. `gamma`-rays

Answer

A

Explanation

The cut-off wavelength `lamda_("min")` corresponds to an electron transferring (approximately) all of its energy to an X-ray photon, thus producing a photon with the greatest possible frequency and least possible wavelength.
From relation, `lamda_("min")=(hc)/(K_(0))=((4.14xx10^(-15))(3xx10^(8)))/(35.0xx10^(3))`
`=3.55xx10^(-11)m=35.5`pm