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 = `E_1` =  ? 

(b) Energy of photon in Greenlight = `E_2` =  ? 

(c) Energy of photon in X-ray = `E_3` =  ? 




>

Solution:


We know that 

E = hf 

where f = `\frac {c}{λ}` So,

E = `\frac {hc}{λ}` -------------(1)



(a) Energy of photon in Radio wave = `E_1` =  ? 

Using equation (1)

`E_1` = `\frac {hc}{λ₁}`

By putting the corresponding values

`E_1` = `\frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ3 ｘ 10⁸ m s⁻¹}{100 m}`

`E_1` = `\frac {19.878 ｘ 10⁻²⁴ m³ kg s⁻² }{100 m}`

`E_1` = 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

`E_1` = `\frac {0.19878 ｘ 10⁻²⁴ }{1.6 ｘ 10⁻¹⁹ }` eV

`E_1` = 1.24 ｘ 10⁻⁸ eV   --------------Ans.1


(b) Energy of photon in Greenlight = `E_2` =  ? 

Using equation (1)

`E_2` = `\frac {hc}{λ₁}`

By putting the corresponding values

`E_2` = `\frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ3 ｘ 10⁸ m s⁻¹}{550ｘ 10⁻⁹ m}`

`E_2` = `\frac {19.878 ｘ 10⁻²⁴ m³ kg s⁻² }{550ｘ 10⁻⁹ m}`

`E_2` = 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

`E_2` = `\frac {0.03614 ｘ 10⁻¹⁷ }{1.6 ｘ 10⁻¹⁹ }` eV

`E_2` = 0.0225 ｘ 10² eV

or

`E_2` = 2.25  eV   --------------Ans.2



(c) Energy of photon in X-ray = `E_3` =  ? 

Using equation (1)

`E_3` = `\frac {hc}{λ₁}`

By putting the corresponding values

`E_3` = `\frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ3 ｘ 10⁸ m s⁻¹}{0.2ｘ 10⁻⁹ m}`

`E_3` = `\frac {19.878 ｘ 10⁻²⁴ m³ kg s⁻² }{0.2ｘ 10⁻⁹ m}`

`E_3` = 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

`E_3` = `\frac {99.39 ｘ 10⁻¹⁷ }{1.6 ｘ 10⁻¹⁹ }` eV

`E_3` = 62.119 ｘ 10² eV

or

`E_3` = 6211.9  eV   --------------Ans.3




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