Question

If a source of power 4 kW produces $ {10^{20}} $ photons/second, the radiation belongs to a part of the spectrum called
(A) $ \gamma - rays $
(B) X – rays
(C) Ultraviolet rays
(D) Microwave.

Answer 1

Hint: The energy of a photon is directly proportional to the frequency of one photon (constant of proportionality is the Planck’s constant). The power of the source determines the rate at which the photons are ejected, hence determines the energy given to one photon.

Formula used: In this solution we will be using the following formula;
 $ E = hf $ where $ E $ is the energy of one photon, $ f $ is the frequency of the photon, and $ h $ is called the Planck’s constant.
 $ P = \dfrac{E}{t} $ where $ P $ is the power rating of the source, is the energy released by the source through a time, and $ t $ is the time.

Complete step by step answer
In the question, a particular amount of photon is said to be ejected per second. By the Planck’s relation, the energy of one of those photons can be given as
 $ {E_p} = hf $ where, $ f $ is the frequency of the photon, and $ h $ is called the Planck’s constant.
Now a particular source is said to have a power of rating 4 kW. The energy released in a particular time is hence given as
 $ E = Pt $ (from $ P = \dfrac{E}{t} $ ) where $ P $ is the power rating of a source, and $ t $ is the time considered.
The energy this source releases is the photons of light, hence the energy released after a particular time is the number of photons released after that same time (at a rate of $ {10^{20}} $ photons/second), hence
 $ E = {E_p} $ hence,
 $ E = nhf $, where $ n $ is the number of photons.
Thus on equating both the values we get,
 $ nhf = Pt $
 $ P = \dfrac{{nhf}}{t} = \dfrac{n}{t}hf $ where $ \dfrac{n}{t} $ is the rate of ejection of photons, hence by insertion of all values and making $ f $ subject of formula
 $ f = \dfrac{{4 \times {{10}^3}}}{{{{10}^{20}} \times 6.63 \times {{10}^{ - 34}}}} = 6.03 \times {10^{16}}Hz $
This is the range of frequency of the X-rays.

Hence, the correct option B.

Note
Alternatively, in the more familiar wavelength format, we have
 $ \lambda = \dfrac{c}{f} $ hence, by inserting known values, we have that
 $ \lambda = \dfrac{{3 \times {{10}^8}}}{{6.03 \times {{10}^{16}}}} = 4.98 \times {10^{ - 9}}m \approx 5nm $.
This is within the range of the wavelength of X-rays.