Find the energy of the photon in
(a) Radio wave of wavelength 100m
(b) Green light of wavelength 550nm
(c) X-ray with wavelength 0.2nm.

[Ans. (a)1.24 ｘ 10⁻⁸ eV, (b) 2.25 eV, (c) 6212 eV]



Given Data:

Let

(a) Radio wave of wavelength = λ₁ = 100 m

(b) Green light of wavelength = λ₂ = 550 nm = 550ｘ 10⁻⁹ m

(c) X-ray with wavelength = λ₃ = 0.2 nm = 0.2ｘ 10⁻⁹ m

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Speed of light = 3 ｘ 10⁸ m s⁻¹


To Find:

(a) Energy of photon in Radio wave = E1 =  ? 

(b) Energy of photon in Greenlight = E2 =  ? 

(c) Energy of photon in X-ray = E3 =  ? 




>

Solution:


We know that 

E = hf 

where f = cλ So,

E = hcλ -------------(1)



(a) Energy of photon in Radio wave = E1 =  ? 

Using equation (1)

E1 = hcλ₁

By putting the corresponding values

E1 = 6.626ｘ10⁻³⁴m²kgs⁻¹ｘ3ｘ10⁸ms⁻¹100m

E1 = 19.878ｘ10⁻²⁴m³kgs⁻²100m

E1 = 0.19878 ｘ 10⁻²⁴ J

by expressing in electron Volts (eV) we have [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ] So, by using the unitary method we have

E1 = 0.19878ｘ10⁻²⁴1.6ｘ10⁻¹⁹ eV

E1 = 1.24 ｘ 10⁻⁸ eV   --------------Ans.1


(b) Energy of photon in Greenlight = E2 =  ? 

Using equation (1)

E2 = `\frac {hc}{λ₁}`

By putting the corresponding values

E2 = 6.626ｘ10⁻³⁴m²kgs⁻¹ｘ3ｘ10⁸ms⁻¹550ｘ10⁻⁹m

E2 = 19.878ｘ10⁻²⁴m³kgs⁻²550ｘ10⁻⁹m

E2 = 0.03614 ｘ 10⁻¹⁷ J

by expressing in electron Volts (eV) we have [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ] So, by using the unitary method we have

E2 = 0.03614ｘ10⁻¹⁷1.6ｘ10⁻¹⁹ eV

E2 = 0.0225 ｘ 10² eV

or

E2 = 2.25  eV   --------------Ans.2



(c) Energy of photon in X-ray = E3 =  ? 

Using equation (1)

E3 = hcλ₁

By putting the corresponding values

E3 = 6.626ｘ10⁻³⁴m²kgs⁻¹ｘ3ｘ10⁸ms⁻¹0.2ｘ10⁻⁹m

E3 = 19.878ｘ10⁻²⁴m³kgs⁻²0.2ｘ10⁻⁹m

E3 = 99.39 ｘ 10⁻¹⁷ J

by expressing in electron Volts (eV) we have [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ] So, by using the unitary method we have

E3 = 99.39ｘ10⁻¹⁷1.6ｘ10⁻¹⁹ eV

E3 = 62.119 ｘ 10² eV

or

E3 = 6211.9  eV   --------------Ans.3




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