复习提纲

概念题

麦克斯韦方程组(Maxwell’s Equations)

∇ × H = J + ∂ D ∂ t \nabla \times H=J+\dfrac{\partial D}{\partial t} ∇×H=J+∂t∂D

∇ × E = − ∂ B ∂ t \nabla \times E=-\dfrac{\partial B}{\partial t} ∇×E=−∂t∂B

∇ ⋅ B = 0 \nabla \cdot B=0 ∇⋅B=0

∇ ⋅ D = ρ \nabla \cdot D= \rho ∇⋅D=ρ
波矢：wave vector

波前：wave front

光强：light intensity

折射率：refractive index
群速度(group velocity)

v g = c N g v_g = \dfrac{c}{N_g} vg=Ngc，其中 N g = 1 − λ d n d λ N_g=1-\lambda \dfrac{dn}{d\lambda} Ng=1−λdλdn
消逝波(evanescent wave)

α 2 = 2 π n 2 λ [ ( n 1 n 2 ) 2 sin ⁡ 2 θ i − 1 ] 1 / 2 \alpha _2=\dfrac{2\pi n_2}{\lambda}[(\dfrac{n_1}{n_2})^2\sin^2 \theta_i-1]^{1/2} α2=λ2πn2[(n2n1)2sin2θi−1]1/2

δ = 1 α 2 \delta = \dfrac{1}{\alpha _2} δ=α21(penetration depth)
全反射(total internal refraction)

θ c = arcsin ⁡ n 2 n 1 \theta _c=\arcsin \dfrac{n_2}{n_1} θc=arcsinn1n2
斯涅尔定律(snell’s law)

n 1 sin ⁡ θ 1 = n 2 sin ⁡ θ 2 n_1 \sin \theta _1 = n_2 \sin \theta_2 n1sinθ1=n2sinθ2
菲涅尔方程(Fresnel’s Equations)

r s = cos ⁡ θ i − [ n 2 − sin ⁡ 2 θ i ] 1 / 2 cos ⁡ θ i + [ n 2 − sin ⁡ 2 θ i ] 1 / 2 r_s=\dfrac{\cos \theta_i-[n^2-\sin^2\theta_i]^{1/2}}{\cos \theta_i+[n^2-\sin^2\theta_i]^{1/2}} rs=cosθi+[n2−sin2θi]1/2cosθi−[n2−sin2θi]1/2

t s = 2 cos ⁡ θ i cos ⁡ θ i + [ n 2 − sin ⁡ 2 θ i ] 1 / 2 t_s=\dfrac{2\cos \theta_i}{\cos \theta_i+[n^2-\sin^2\theta_i]^{1/2}} ts=cosθi+[n2−sin2θi]1/22cosθi

r p = [ n 2 − sin ⁡ 2 θ i ] 1 / 2 − n 2 cos ⁡ θ i [ n 2 − sin ⁡ 2 θ i ] 1 / 2 + n 2 cos ⁡ θ i r_p=\dfrac{[n^2-\sin^2\theta_i]^{1/2}-n^2\cos \theta_i}{[n^2-\sin^2\theta_i]^{1/2}+n^2\cos \theta_i} rp=[n2−sin2θi]1/2+n2cosθi[n2−sin2θi]1/2−n2cosθi

t p = 2 n cos ⁡ θ i [ n 2 − sin ⁡ 2 θ i ] 1 / 2 + n 2 cos ⁡ θ i t_p=\dfrac{2n\cos \theta_i}{[n^2-\sin^2\theta_i]^{1/2}+n^2\cos \theta_i} tp=[n2−sin2θi]1/2+n2cosθi2ncosθi

当 θ i = 0 \theta_i = 0 θi=0时：

r s = r p = n 1 − n 2 n 1 + n 2 r_s=r_p=\dfrac{n_1-n_2}{n_1+n_2} rs=rp=n1+n2n1−n2

t s = t p = 2 n 1 n 1 + n 2 t_s=t_p=\dfrac{2n_1}{n_1+n_2} ts=tp=n1+n22n1

R = R s = R p = ( n 1 − n 2 n 1 + n 2 ) 2 R=R_s=R_p=(\dfrac{n_1-n_2}{n_1+n_2})^2 R=Rs=Rp=(n1+n2n1−n2)2

T = T s = T p = 4 n 1 n 2 ( n 1 + n 2 ) 2 T=T_s=T_p=\dfrac{4n_1n_2}{(n_1+n_2)^2} T=Ts=Tp=(n1+n2)24n1n2
Brewster’s angle

θ p = arctan ⁡ n 2 n 1 \theta _p = \arctan \dfrac{n_2}{n_1} θp=arctann1n2
减反膜原理

2 π n 2 λ ⋅ 2 d = m π \dfrac{2\pi n_2}{\lambda}\cdot2d=m\pi λ2πn2⋅2d=mπ，所以 d = m ⋅ λ 4 n 2 d=m\cdot \dfrac{\lambda}{4n_2} d=m⋅4n2λ

R m i n = ( n 2 2 − n 1 n 3 n 2 2 + n 1 n 3 ) 2 R_{min}=(\dfrac{n_2^2-n_1n_3}{n_2^2+n_1n_3})^2 Rmin=(n22+n1n3n22−n1n3)2，当 n 2 = n 1 n 3 n_2=\sqrt{n_1n_3} n2=n1n3 时， R m i n = 0 R_{min}=0 Rmin=0
杨氏双缝

极大值位置： x = m ⋅ D λ d x=m\cdot D\dfrac{\lambda}{d} x=m⋅Ddλ

极小值位置： x = ( m + 1 2 ) ⋅ D λ d x=(m+\dfrac{1}{2})\cdot D\dfrac{\lambda}{d} x=(m+21)⋅Ddλ

Δ x = D ⋅ λ d \Delta x=D\cdot\dfrac{\lambda}{d} Δx=D⋅dλ
耗尽区：depletion region

载流子：carrier

内建电场：built-in electric field

同质结：homojunction

异质结：heterojunction
质量作用定律(mass action law)

n p = n i 2 = N c N v exp ⁡ ( − E g k B T ) np=n_i^2=N_cN_v\exp(-\dfrac{E_g}{k_BT}) np=ni2=NcNvexp(−kBTEg)

N c = 2 ( 2 π m e ∗ k B T ) 3 / 2 h 3 N_c=\dfrac{2(2\pi m_e^*k_BT)^{3/2}}{h^3} Nc=h32(2πme∗kBT)3/2

N v = 2 ( 2 π m h ∗ k B T ) 3 / 2 h 3 N_v=\dfrac{2(2\pi m_h^*k_BT)^{3/2}}{h^3} Nv=h32(2πmh∗kBT)3/2
费米能(Fermi Energy)

E F i = E v + 1 2 E g − 1 2 k B T ln ⁡ ( N c N v ) = E v + 1 2 E g − 3 4 k B T ln ⁡ ( m e ∗ m h ∗ ) E_{F_i}=E_v+\dfrac{1}{2}E_g-\dfrac{1}{2}k_BT\ln(\dfrac{N_c}{N_v})=E_v+\dfrac{1}{2}E_g-\dfrac{3}{4}k_BT\ln(\dfrac{m_e^*}{m_h^*}) EFi=Ev+21Eg−21kBTln(NvNc)=Ev+21Eg−43kBTln(mh∗me∗)
LED发光原理：自发辐射
- 同质节LED：ｎ区重掺杂，电子到ｐ区复合；缺点：光子被吸收
  
  h ν 0 = E g + 1 2 k B T h\nu _0 =E_g+\dfrac{1}{2}k_BT hν0=Eg+21kBT
- 异质结LED：限制载流子，减少吸收
- QWLED：
- LED效率：
  
  η I Q E = τ r − 1 τ r − 1 + τ n r − 1 \eta _{IQE}=\dfrac{\tau_r^{-1}}{\tau_r^{-1}+\tau_{nr}^{-1}} ηIQE=τr−1+τnr−1τr−1
  
  η E E = P o ( i n t ) P o \eta _{EE}=\dfrac{P_{o(int)}}{P_o} ηEE=PoPo(int)
  
  η E Q E = P o / h ν I / e \eta _{EQE}=\dfrac{P_o/h\nu}{I/e} ηEQE=I/ePo/hν
  
  η P C E = P o I V \eta _{PCE}=\dfrac{P_o}{IV} ηPCE=IVPo
  
  η L E = Φ ν I V \eta _{LE}=\dfrac{\Phi_{\nu}}{IV} ηLE=IVΦν
  
  Φ ν = P o × ( 633 l m W − 1 ) × V ( λ ) \Phi _{\nu}=P_o\times(633lmW^{-1})\times V(\lambda) Φν=Po×(633lmW−1)×V(λ)
- LED types
  
  SLED:surface emitting
  
  ELED:edge emitting
- 白光：
  
  光致发光（光照发光）：蓝光照射黄光LED
形成激光的必要条件
- 受激辐射(stimulated emission)
- 粒子数反转(population inversion)
- 谐振腔(resonant cavity)
激光的几个特征
- 同相位(phase)
- 同传播方向(direction of propagation)
- 同频率(frequency)
- 同极化方向(direction of polarization)
红宝石激光器发光原理

红宝石激光器由红宝石棒、脉冲氙灯、聚光器和谐振腔组成

可看作是一个三能级系统
量子阱

材料构成:窄带隙半导体材料和宽带隙半导体材料

结构：宽带隙势垒层夹着一个窄带隙超薄层

能级结构： E n = E c + h 2 n 2 8 m e ∗ d 2 E_n=E_c+\dfrac{h^2n^2}{8m_e^*d^2} En=Ec+8me∗d2h2n2
光探测器的输入输出

计算题

课本例题

1.1.1(P26)

Consider a He-Ne laser beam at 633 nm with a spot size of 1 mm. Assuming a Gaussian beam, what is the divergence of the beam? What are the Rayleigh range and the beam width at 25 m?

Solution:

divergence: 2 θ = 4 λ π ( 2 w o ) 2\theta =\dfrac{4\lambda}{\pi (2w _o)} 2θ=π(2wo)4λ

Rayleigh range: z 0 = π w 0 2 λ z_0 =\dfrac{\pi w_0^2}{\lambda} z0=λπw02

beam width: 2 w = 2 w 0 [ 1 + ( z λ π w 0 2 ) 2 ] 1 / 2 2w=2w_0[1+(\dfrac{z\lambda}{\pi w _0^2})^2]^{1/2} 2w=2w0[1+(πw02zλ)2]1/2
1.4.2(P37)

Consider a 5 mW He-Ne laser that is operating at 633 nm, and has a spot size of 1 mm. Find the maximum irradiance of the beam and the axial(maximum) irradiance at 25 m from the laser.

Solution:

maximum irradiance of the beam: I 0 = 2 P 0 π w 0 2 I_0=\dfrac{2P_0}{\pi w_0^2} I0=πw022P0

axial irradiance: I a x i s = I 0 ( z 0 z ) 2 = I 0 ( π w 0 2 λ z ) 2 I_{axis}=I_0(\dfrac{z_0}z)^2=I_0(\dfrac{\pi w_0^2}{\lambda z})^2 Iaxis=I0(zz0)2=I0(λzπw02)2
1.11.1(P73)

Consider a Farbry-Perot optical cavity made of a semiconductor material with mirrors at its ends. The length of the semiconductor , and hense the cavity, is 250 μ \mu μm and mirrors at the ends have a reflectance of 0.90. Calculate the cavity mode nearest to the free-space wavelength of 1310 nm. Calculate the separation of the modes, finesse, spectral width of each mode in width of each mode in frequency and wavelength, and the Q-factor.

Solution:

nearest mode: m = 2 n L λ = 1374.05 m=\dfrac{2nL}\lambda=1374.05 m=λ2nL=1374.05

λ m = 2 n L m \lambda_m=\dfrac{2nL}m λm=m2nL

separation: Δ ν m = ν f = c 2 n L \Delta \nu_m=\nu _f=\dfrac{c}{2nL} Δνm=νf=2nLc

finesse: F = π R 1 − R F=\dfrac{\pi\sqrt{R}}{1-R} F=1−RπR

width: δ ν m = ν f F , δ λ m = δ ( c ν m ) = ∣ − c ν m 2 ∣ δ ν m \delta \nu _m=\dfrac{\nu _f}{F}, \delta\lambda _m=\delta(\dfrac c{\nu _m})=\left|-\dfrac{c}{\nu _m^2}\right|\delta\nu _m δνm=Fνf,δλm=δ(νmc)=∣∣∣∣−νm2c∣∣∣∣δνm

Q-factor: Q = m F Q=mF Q=mF
3.3.1(P209)Fermi levels in semiconductors

An n-type Si wafer has been doped uniformly with 1 0 16 10^{16} 1016 phosphorus atoms c m − 3 cm^{-3} cm−3. Calculate the position of the Fermi energy with respect to the Fermi energy E F i E_{F_i} EFi in intrisic Si. The above n-type Si sample is further doped with 2 × 1 0 17 2\times 10^{17} 2×1017 boron atoms c m − 3 cm^{-3} cm−3. Calculate the position of the Fermi energy with respect to the Fermi energy E F i E_{F_i} EFi in intrisic Si at the room temperature(300K), and hence with respect to the Fermi energy in the n-type case above.

Solution:

n = N c exp ⁡ [ − E c − E F n k B T ] n=N_c\exp[-\dfrac{E_c-E_{F_n}}{k_BT}] n=Ncexp[−kBTEc−EFn]

n i = N c exp ⁡ [ − E c − E F i k B T ] n_i=N_c\exp[-\dfrac{E_c-E_{F_i}}{k_BT}] ni=Ncexp[−kBTEc−EFi]

n n i = exp ⁡ [ E F n − E F i k B T ] \dfrac{n}{n_i}=\exp[\dfrac{E_{F_n}-E_{F_i}}{k_BT}] nin=exp[kBTEFn−EFi]

∴ E F n − E F i = k B T ln ⁡ ( n / n i ) \therefore E_{F_n}-E_{F_i}=k_BT\ln(n/n_i) ∴EFn−EFi=kBTln(n/ni)

similarly: ∴ E F p − E F i = − k B T ln ⁡ ( p / n i ) \therefore E_{F_p}-E_{F_i}=-k_BT\ln(p/n_i) ∴EFp−EFi=−kBTln(p/ni)

∴ E F n − E F p = k B T ln ⁡ ( n / p ) \therefore E_{F_n}-E_{F_p}=k_BT\ln(n/p) ∴EFn−EFp=kBTln(n/p)
3.9.1(P237)The built-in voltage from the band diagram

e V o = Φ p − Φ n = E F n − E F p ( b e f o r e c o n t a c t ) eV_o=\Phi _p-\Phi _n=E_{F_n}-E_{F_p}(before\ contact) eVo=Φp−Φn=EFn−EFp(before contact)

n = N c exp ⁡ ( − E c − E F n k B T ) n=N_c\exp(-\dfrac{E_c-E_{F_n}}{k_BT}) n=Ncexp(−kBTEc−EFn)

p = N v exp ⁡ ( − E F p − E v k B T ) p=N_v\exp(-\dfrac{E_{F_p}-E_v}{k_BT}) p=Nvexp(−kBTEFp−Ev)

∴ n p = N a N d = N c N v exp ⁡ ( − E c − E v + E F p − E F n k B T ) = n i 2 exp ⁡ ( F n − E F p k B T ) \therefore np=N_aN_d=N_cN_v\exp(-\dfrac{E_c-E_v+E_{F_p}-E_{F_n}}{k_BT})=n_i^2\exp(\dfrac{{F_n}-E_{F_p}}{k_BT}) ∴np=NaNd=NcNvexp(−kBTEc−Ev+EFp−EFn)=ni2exp(kBTFn−EFp)

e V o = k B T ln ⁡ [ ( N a N d ) / n i 2 ] eV_o=k_BT\ln[(N_aN_d)/n_i^2] eVo=kBTln[(NaNd)/ni2]
4.10.1(P336)Modes in a semiconductor laser and the optical cavity length

Consider an AlGaAs-based heterostructure laser diode that has an optial cavity of length 200 μ m \mu m μm. The peak radiation is at 870 nm and the refractive index of GaAs is about 3.6. What is the mode integer m of the peak radiation and the separation between the modes of cavity? If the optial gain vs. wavelength characteristics has a FWHM wavelength width of about 6 nm, how many modes are there within this bandwidth?How many modes are there if the cavity length is 20 μ m \mu m μm?

Solution:

m = 2 n L λ m=\dfrac{2nL}{\lambda} m=λ2nL

Δ λ m = 2 n L m − 2 n L m + 1 = 2 n L m 2 = λ 2 2 n L \Delta \lambda _m=\dfrac{2nL}m-\dfrac{2nL}{m+1}=\dfrac{2nL}{m^2}=\dfrac{\lambda ^2}{2nL} Δλm=m2nL−m+12nL=m22nL=2nLλ2

number of modes: Δ λ 1 / 2 Δ λ m \dfrac{\Delta\lambda_{1/2}}{\Delta\lambda_m} ΔλmΔλ1/2
4.11.1(P339)A GaAs quantum well

Consider a very thin GaAs quantum well sandwiched between two wider bandgap semiconductor layers of AlGaAs. The QW depth from E c E_c Ec and E v E_v Ev are approximately 0.28 eV and 0.16 eV, respectively. Effective mass m e ∗ m_e^* me∗ of a conduction electron in GaAs is approximately 0.07 m e m_e me, where m e m_e me is the electron mass in vacuum. Calculate the first two electron energy levels for a quantum well of thickness 10 nm. What is the hole energy in the QW above E v E_v Ev of GaAs if the hole effective mass m h ∗ ≈ 0.50 m e m_h^*\approx0.50m_e mh∗≈0.50me? What is the change oin the emission wavelength with respect to bulk GaAs, for which E g = 1.42 e V E_g=1.42eV Eg=1.42eV? Assume infinite QW depths for the calculations.

Solution:

ε n = E n − E c = h 2 n 2 8 m e ∗ d 2 \varepsilon_n=E_n-E_c=\dfrac{h^2n^2}{8m_e^*d^2} εn=En−Ec=8me∗d2h2n2

ε n ′ = E v − E n = h 2 n ′ 2 8 m h ∗ d 2 \varepsilon_{n\prime}=E_v-E_n=\dfrac{h^2n\prime^2}{8m_h^*d^2} εn′=Ev−En=8mh∗d2h2n′2

λ g = h c E g \lambda_g=\dfrac{hc}{E_g} λg=Eghc

λ Q W = h c E g + ε 1 + ε 1 ′ \lambda_{QW}=\dfrac{hc}{E_g+\varepsilon_1+\varepsilon_1^\prime} λQW=Eg+ε1+ε1′hc
4.12.1(P346)Laser output wavelength wariation with temperature

The refractive index n of GaAs is approximately 3.6 and it has a temperature dependence d n / d T ≈ 2.0 × 1 0 − 4 K − 1 dn/dT\approx2.0\times10^{-4}K^{-1} dn/dT≈2.0×10−4K−1. Estimate the change in th emitted wavelength at around 870 nm per degree change in the temperature for a given mode.

Solution:

d λ m d T = d d T [ 2 m n L ] = 2 L m d n d T = λ m n d n d T \dfrac{d\lambda_m}{dT}=\dfrac{d}{dT}[\dfrac2mnL]=\dfrac{2L}m\dfrac{dn}{dT}=\dfrac{\lambda_m}n\dfrac{dn}{dT} dTdλm=dTd[m2nL]=m2LdTdn=nλmdTdn

课后习题

1.1Maxwell’s wave equation and plane waves

∂ 2 E ∂ x 2 + ∂ 2 E ∂ y 2 + ∂ 2 E ∂ z 2 − ε 0 ε r μ 0 ∂ 2 E ∂ t 2 = 0 \dfrac{\partial^2E}{\partial x^2}+\dfrac{\partial^2E}{\partial y^2}+\dfrac{\partial^2E}{\partial z^2}-\varepsilon_0\varepsilon_r\mu_0\dfrac{\partial^2E}{\partial t^2}=0 ∂x2∂2E+∂y2∂2E+∂z2∂2E−ε0εrμ0∂t2∂2E=0
3.6Electrons in GaAs

Given that the electron effective mass m e ∗ m_e^* me∗ for the GaAs is 0.067 m e m_e me, calculate the thermal velocity of electrons in the CB of a nondegenately doped GaAs at room temperture. If μ e \mu_e μe is the drift mobility of the electrons and τ e \tau_e τe the mean free time between electron scaterring everents and if μ e = e τ e / m e ∗ \mu_e=e\tau_e/m_e^* μe=eτe/me∗, calculate τ e \tau_e τe, given μ e = 8500 c m 2 V − 1 s − 1 \mu_e=8500cm^2V^{-1}s^{-1} μe=8500cm2V−1s−1. Calculate the drift velocity v d = μ e E v_d=\mu_eE vd=μeE of the CB electrons in an applied field E of 1 0 5 V m − 1 10^5Vm^{-1} 105Vm−1.

Solution:

1 2 m e ∗ v t h 2 = 3 2 k B T \dfrac{1}2m_e^*v_{th}^2=\dfrac32k_BT 21me∗vth2=23kBT

τ e = m e ∗ μ e e \tau_e=\dfrac{m_e^*\mu_e}e τe=eme∗μe

v d = μ e E v_d=\mu_eE vd=μeE
3.9Compensation doping in n-type Si

An n-type Si sample has been doped with 1 0 16 10^{16} 1016 phosphorus atoms c m − 3 cm^{-3} cm−3. ( a )What are the electron and hole concentrations?( b )Calculate the room temperature conductivety of the sample. ( c )Where is the Fermi level with respect to E F i E_{F_i} EFi? ( d )If we now dope the crystal with 1 0 17 10^{17} 1017 boron acceptors, what will be the electron and hole concentrations? ( e )Where is the Fermi level with respect to E F i E_{F_i} EFi.

Solution:

( a ) n = N d , p = n i 2 / n n=N_d,\ p=n_i^2/n n=Nd, p=ni2/n

( b ) σ = e n μ e + e p μ h \sigma=en\mu_e+ep\mu_h σ=enμe+epμh

( c ) n = N c exp ⁡ ( − E − E F n k B T ) n=N_c\exp(-\dfrac{E_-E_{F_n}}{k_BT}) n=Ncexp(−kBTE−EFn)

n i = N c exp ⁡ ( − E c − E F i k B T ) n_i=N_c\exp(-\dfrac{E_c-E_{F_i}}{k_BT}) ni=Ncexp(−kBTEc−EFi)

n / n i = exp ⁡ ( E F n − E F i k B T ) n/n_i=\exp(\dfrac{E_{F_n}-E_{F_i}}{k_BT}) n/ni=exp(kBTEFn−EFi)

E F n − E F i = k B T ln ⁡ ( N d / n i ) E_{F_n}-E_{F_i}=k_BT\ln(N_d/n_i) EFn−EFi=kBTln(Nd/ni)

( d ) p = N a − N d , n = n i 2 / p p=N_a-N_d,\ n=n_i^2/p p=Na−Nd, n=ni2/p

( e ) E F p − E F i = k B T ln ⁡ ( n i / p ) E_{F_p}-E_{F_i}=k_BT\ln(n_i/p) EFp−EFi=kBTln(ni/p)
3.29LED efficients

A partiaclar 890 nm IR LED for use instrumentation has an AlGaAs chip. The cative region has been doped p-type with 4 × 1 0 17 c m − 3 4\times10^{17}cm^{-3} 4×1017cm−3 of acceptors and the nonradiative lifetime is about 60 ns. At a forward current of 50 mA, the voltage across it is 1.4 V, and the emitted optical power is 10 mW. Calculate the PCE, IQE, EQE, and estimate the light eatration ratio. For AlGaAs, B ≈ 1 × 1 0 16 m 3 s − 1 B\approx1\times10^{16}m^3s^{-1} B≈1×1016m3s−1.

Solution:

τ r = 1 B N a \tau_r=\dfrac{1}{BN_a} τr=BNa1

η I Q E = τ r − 1 τ r − 1 + τ n r − 1 \eta_{IQE}=\dfrac{\tau_r^{-1}}{\tau_r^{-1}+\tau_{nr}^{-1}} ηIQE=τr−1+τnr−1τr−1

η P C E = P o I V \eta_{PCE}=\dfrac{P_o}{IV} ηPCE=IVPo

η E Q E = P o / h ν I / E \eta_{EQE}=\dfrac{P_o/h\nu}{I/E} ηEQE=I/EPo/hν

η E E = η I W Q / η E W Q \eta_{EE}=\eta_{IWQ}/\eta_{EWQ} ηEE=ηIWQ/ηEWQ
4.13 Photon concentration in a gas laser

The Ar ion laser has a strong lasing emission at 488 nm. The laser tube is 1 m in length, and the bore diameter is 3 mm. The output power is 1 W. Assume that most of the out put power is in the 488 nm emission. Assume that the tube end has a transittance T of 0.1. Calcultate the photon output flow, photon flux, and estimate the order of magnitude of the steady state photon concentration in the tube.

Solution:

P h o t o n F l o w = P o h ν Photon\ Flow=\dfrac{P_o}{h\nu} Photon Flow=hνPo

Γ p h = P o A h ν \Gamma_{ph}=\dfrac{P_o}{Ah\nu} Γph=AhνPo

P o h ν = T n p h A L Δ t \dfrac{P_o}{h\nu}=T\dfrac{n_{ph}AL}{\Delta t} hνPo=TΔtnphAL

Δ t = 2 L / c \Delta t=2L/c Δt=2L/c

n p h = P o T A c h ν n_{ph}=\dfrac{P_o}{TAch\nu} nph=TAchνPo
constant

k B = 1.38 × 1 0 − 23 J / K k_B=1.38\times10^{-23}J/K kB=1.38×10−23J/K

h = 6.626 × 1 0 − 34 J ⋅ s h=6.626\times10^{-34}J\cdot s h=6.626×10−34J⋅s

m e = 9.10956 × 1 0 − 31 k g m_e=9.10956\times10^{-31}kg me=9.10956×10−31kg

e = 1.602 × 1 0 − 19 C e=1.602\times10^{-19}C e=1.602×10−19C

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