## 6 Common Problems of Photons

- Category: High School
- Written by fisikastudycenter

physics.fisikastudycenter.com,- Learning energy of photons, the number of photons and its relationships with radiated power and radiation intensity. These topics are discussed at grade 12 high school. Follow these questions and answers below to get more understanding.

**Equations**

**Energy of a Photon**

E = hf

E = h( ^{c}/_{λ} )

**Energy of n Photons**

E = nhf

E = nh( ^{c}/_{λ} )

**Conversion factors**

1 electron volt = 1 eV = 1,6 x 10^{−19} joule

1 angstrom = 1 Å = 10^{−10} meter

1 nanometer = 1 nm = 10^{−9} meter

Power → Energy per second

Intensity → Energy per second per m^{2}

**Problem ****1**

Find the consisted energy of light with a wavelength of 6600 Å if the speed of light is 3 x 10^{8} m/sec and the Planck's constant h = 6,6 x 10^{−34} Js!

**Answer**

E = h(^{c}/_{λ})

E = (6,6 x 10^{−34} )( ^{3 x 108}/_{6600 x 10−10} ) = 3 x 10^{−19} joule

**Problem 2**

A monochromatic lamp of 100 watt in power radiates a wavelength of 5,5**.**10^{−7} m. Find the number of radiated photon per second!

A. 2,8 x 10^{22} /s

B. 2,0 x 10^{22} /s

C. 2,6 x 10^{20} /s

D. 2,8 x 10^{20} /s

E. 2,0 x 10^{20} /s

**Answer**

Data :

P = 100 watt → Energy per second is 100 joule.

Energy of a photon

E = h(^{c}/_{λ})

E = (6,6 x 10^{−34} )( ^{3 x 108}/_{5,5 x 10−7} ) joule

Number of photon (n)

n = 100 joule : [ (6,6 x 10^{−34} )( ^{3 x 108}/_{5,5 x 10−7} ) joule] = 2,8 x 10^{20} foton. ** **

**Problem**** 3**

The intensity of radiation on the wall received from a stove is 66,3 W.m^{−2}. Assume the stove as a black body and radiates a wavelength of 600 nm, then find the number of photons reach the wall per second per unit area. Use h = 6,63 x 10^{− 34} J.s, c = 3 x 10^{8} m.s^{− 1}!

**Answer**

Data :

I = 66,3 W.m^{−2} → The received energy per second per m^{2}

energy of a photon

E = h(^{c}/_{λ})

E = (6,63 x 10^{−34} )( ^{3 x 108}/_{600 x 10−9} ) joule

the number of photon:

n = 66,3 joule : [ (6,63 x 10^{−34} )( ^{3 x 108}/_{600 x 10−9} ) joule] = 2 x 10^{20} foton

**Problem 4**

Find the ratio of energy consisted by a light of 6000 Å of wavelenght with a light of 4000 Å of wavelenght!

**Answer**

Data :

λ_{1} = 6000 Å

λ_{2} = 4000 Å

E = h(^{c}/_{λ})

^{E1}/_{E2} = λ_{2} : λ_{1} = 4000 : 6000 = 2 : 3

**Problem 5**

Photon energy of a gamma ray is 10^{8} eV. If h = 6,6 x 10^{−34} Js and speed of light c = 3 x 10^{8} m/s, find the wavelength show the answer in angstom!

**Answer**

Data :

E = 10^{8} eV = 10^{8} x (1,6 x 10^{−19}) joule = 1,6 x 10^{−11} joule

h = 6,6 x 10^{−34} Js

c = 3 x 10^{8} m/s

λ = ...?

λ = ^{hc} / _{E}

λ = ^{( 6,6 x 10−34)(3 x 108)} / _{(1,6 x 10−11)}

λ = 12,375 x 10^{−15} meter =12,375 x 10^{−15} x 10^{10} Å = 12,375 x 10^{−5} Å

**Problem 6**

A lamp of 132 W/220 V is used at a voltage of V. The magnitude of radiated wavelength is 628 nm. If the radiated power of the lamp spreads around uniformly, find the number of photons at a distance of 2.5 meter from the lamp, use h = 6,6.10^{−34} J s!

**Answer**

First, determine the power of lamp with its specification of 132 W/220 V when used at a voltage of 110 V :

P_{2} =(^{V2}/_{V1})^{2} x P_{1}

P_{2} =(^{110}/_{220})^{2} x 132 watt = 33 watt

Intensity at 2,5 meter :

I = (^{P}/_{A})

I = (^{P}/_{4π r2})

I = (^{33}/_{4π (2,5)2}) = 0,42 watt/m^{2}

0,42 watt/m^{2} → energy per second per unit of area is 0,42 joule.

The number of photon (n) :

n = 0,42 : (^{hc}/_{λ}) = [ 0,42 ] **:** [ ^{( 6,6 x 10−34 )( 3 x 108 )}/_{( 628 x 10−9 )} ] = ( 0,42 ) **:** (3,15 x 10^{−19} )

n = 1,33 x 10^{18} foton