For wave intensity 1 W/m2 and wavelength 1 µm in vaccum, find the number of photons. What is the energy of photon of wavelength 1 mm? Compare this with the characteristic thermal energy at 3000 K.
Answers
Answer:
Hey Mate..!
Explanation:
eV(Ionisation energy)=133.24×10
−16
J
We use the formula
eV=hf (Eq 1)
where eV is the ionisation energy
h is the Planck's constant=6.62×10
−34
f is the frequency f=
λ
C
Where C is the speed of light C=3×10
8
msec
−1
λ is the wavelength
Putting the value of f in equation 1
eV=h
λ
C
133.24×10
−16
=
λ
6.62×10
−34
×3×10
8
λ=1.49×10
−11
m
For the second question
K=hv−hv
0
(eq 1)
where h is the Planck's constant and v
0
is the threshold frequency and hv
0
is the work function and K is the kinetic energy
v=
λ
C
Putting the values in euation 1
1=
λ
hC
−hv
0
(eq 2)
4=
λ
3hC
−hv
0
(eq 3)
eq 2 is multiplied by −3 then adding eq 3 from eq2
−3=
λ
−3hC
+3hv
0
4=
λ
3hC
−hv
0
4−3=(3−1)hv
0
hv
0
=0.5eV
0
Answer:
a) Number of photons in vacuum = 5.03×10¹⁸
b) Energy of photon of wavelength 1mm = 19.878×10⁻²³ joule
Explanation:
Given: intensity of wave is = 1W/m²
wavelength in vacuum= 1μm
To calculate number of photon :
number of photon in vacuum = energy of each photon
b (E) = hν
E = hc/λ
h = 6.626×10⁻³⁴ js
c =3×10⁸ m/s²
λ= 1×10⁻⁶ m
E = (6.626×10 ⁻³⁴)(3×10⁸)/(1×10⁻⁶)
E = 19.88×10 ⁻²⁰
total energy = 1w
number of photon = 1/(19.88×10⁻²⁰)
number of photon in vacuum = 5.03×10¹⁸
Energy of photon of λ = 1mm :
λ= 1mm = 1×10⁻³ m
E= hc/λ
E = (6.626×10⁻³⁴)(3×10⁸)/(1×10⁻³)
E = 19.87×10⁻²³ j
Thermal energy:
E = kT
Boltzmann constant (k)= 1.38×10⁻²³
temperature (T)= 3000K
E =(1.38 ×10⁻²³)(3000)
E = 4140×10⁻²³ j