Solveeit Logo
Login

Question

Question: Find the energy of photon in each of the following: (A). Microwaves of wavelength \[1.5\;{\text{cm...

Find the energy of photon in each of the following:
(A). Microwaves of wavelength 1.5  cm1.5\;{\text{cm}}.
(B). Red light of wavelength 660  nm660\;{\text{nm}}.
(C). Radio waves of frequency 96  MHz96\;{\text{MHz}}
(D). X−{\text{X}} - rays of wavelength 0.17  nm0.17\;{\text{nm}}.

Explanation

Solution

In this question, first obtain the expression for the photon energy in terms of Planck constant, the speed of the light, and the wavelength. Use the constant values of Plank Constant and the speed of the light in the expression to calculate the energy.

Complete step by step answer:

(A)
As we know that the photons are the electrically neutral particles which have no mass, but they have the energy. The SI unit of energy is Joule. The photon energy is calculated by the formula given below.
E=hcλE = \dfrac{{hc}}{\lambda }
Where, hh is the Planks universal constant, the value of Planck's constant is 6.62×10−34  Js6.62 \times {10^{ - 34}}\;{\text{Js}} and cc is the speed of the light is 3×108  m/sec3 \times {10^8}\;{\text{m/sec}}.

First, we convert the wavelength from centimeter to meter,
λ=1.5  cm(1  m100  cm)\lambda = 1.5\;{\text{cm}}\left( {\dfrac{{1\;{\text{m}}}}{{100\;{\text{cm}}}}} \right)
λ=0.015  m\lambda = 0.015\;{\text{m}}

Substitute the value of wavelength in photon energy formula as,
E=6.62×10−34×3×1080.015E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.015}}
Simplify, the above equation to obtain the energy as,
E=1324×10−26  JE = 1324 \times {10^{ - 26}}\;{\text{J}}

Therefore, the energy of the photon is 1324×10−26  J1324 \times {10^{ - 26}}\;{\text{J}}.

(B).
In this question, the wavelength of the red light is 660  nm660\;{\text{nm}}.

First, we convert the wavelength of red light from nanometre to meter,
λ=660  nm\lambda = 660\;{\text{nm}}
λ=660×10−9  m\lambda = 660 \times {10^{ - 9}}\;{\text{m}}

The photon energy is calculated by the formula given below.
E=hcλE = \dfrac{{hc}}{\lambda }

Now, we substitute the values as,
E=6.62×10−34  Js×3×108  m/sec660×10−9  mE = \dfrac{{6.62 \times {{10}^{ - 34}}\;{\text{Js}} \times 3 \times {{10}^8}\;{\text{m/sec}}}}{{660 \times {{10}^{ - 9}}\;{\text{m}}}}
By simplifying the above equation, we get,
E=0.03×10−17  JE = 0.03 \times {10^{ - 17}}\;{\text{J}}
Therefore, the energy of the photon is 0.03×10−17  J0.03 \times {10^{ - 17}}\;{\text{J}}.

(C)
In this question, the radio wave frequency is 96  MHz96\;{\text{MHz}}.

Convert the frequency into Hertz (Hz)\left( {{\text{Hz}}} \right),
v=96  MHzv = 96\;{\text{MHz}}
v=96×106  Hzv = 96 \times {10^6}\;{\text{Hz}}

The photon energy is calculated by the formula given below.
E=hvE = hv

Substitute the values,
E=(6.62×10−34)(96×106)E = \left( {6.62 \times {{10}^{ - 34}}} \right)\left( {96 \times {{10}^6}} \right)
By simplifying the above equation,
E=635.52×10−28  JE = 635.52 \times {10^{ - 28}}\;{\text{J}}

Therefore, the energy of the photon is 635.52×10−28  J635.52 \times {10^{ - 28}}\;{\text{J}}.

(D).
We are given X−X - ray of wavelength 0.17  nm0.17\;{\text{nm}}
Convert the wavelength of X−X - ray from nanometre to meter,
λ=0.17  nm\lambda = 0.17\;{\text{nm}}
λ=0.17×10−9  m\lambda = 0.17 \times {10^{ - 9}}\;{\text{m}}

The photon energy is calculated by the formula given below.
E=hcλE = \dfrac{{hc}}{\lambda }

Now, substitute the given values as,
E=6.62×10−34×3×1080.17×10−9E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.17 \times {{10}^{ - 9}}}}
By simplifying above equation,
E=116.82×10−17  JE = {\text{116}}{\text{.82}} \times {10^{ - 17}}\;{\text{J}}

Therefore, the energy of photon is 116.82×10−17  J{\text{116}}{\text{.82}} \times {10^{ - 17}}\;{\text{J}}.

Note: We know that the frequency is the number of oscillations per unit time and it is the ratio of the speed of light and the wavelength and it is the reciprocal of the time period that is the time taken to complete one oscillation.