If the energy of a photon of sodium light ( `lambda`=589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair
A. 1.1 eV
B. 2.1 eV
C. 3.2 eV
D. 1.5 eV
Correct Answer - B
Using ,`E=E_g=(hc)/lambda`
`=(6.6xx10^(-34)xx3xx10^8)/(589xx10^(-9))J`
`=(6.6xx10^(-34)xx3xx10^8)/(589xx10^(-9)xx1.6xx10^(-19))`eV =2.1 eV
Correct Answer - D
The de Brogli wavelength
`h=(h)/(mv)rArrv=(h)/(mlambda)`
Energy of photon
`E_(p)=(hc)/(lambda)(since lambda is same)`
Energy of Photon
Kinetic energy of photon `=(E_(p))/(E_(e))=(hc//lambda)/((1)/(2)1mu^(2))=(2hc)/(lambdamv^(2))`
substituting value of V from Eq (i)...
Correct Answer - A
Energy gap, `E_g=(hc)/lambda, lambda=(hc)/E_g`
Here energy gap=5.50 eV
Take hc=1240 eV nm
`therefore lambda=(1240 eV nm) /(5.5 eV)=226 nm`
Correct Answer - C
If there is some gap between the conduction band and the valance band electrons in the valance band all remain bound and no free electrons in conduction...
Correct Answer - C
The spectral range of visible light is from about 3 eV to 1.8 eV. Therefore the semiconductor used for fabrication of visible LEDs must at least have...
Correct Answer - C
(c) According to formula =, `E=(hc)/(lambda)(v=(c)/(lambda))`
Energy E=hv
`3.03xx10^(-19)=(hc)/(lambda)`
`lambda=(6.63xx10^(-34)xx3.0xx10^(8))/(3.03xx10^(-19))`
`=6.56xx10^(-7) m`
`=6.56xx10^(-7)xx10^(9)nm`
=`6.56xx10^(2) nm`
`=656 nm`
Let `lambda_(0)` is the threshold wavelength. The work function is `phi = (hc)/(lambda_(0))`
Now, by photoelectric equation `ex = (hc)/(lambda) - (hc)/(lambda_(0))` ……(i) `(ex)/(n+1) = (hc)/(nlambda) - (hc)/(lambda_(0))`……..(ii)
From (i)...
We have `[(lambda^(2)-2lambda+1,lambda-2),(1-lambda^(2)+3lambda,1-lambda^(2))]=Alambda^(2)+Blambda+C`
Putting `lambda=0`, we get
`C=[(1,-2),(1,1)]`
Putting `lambda=1`, we get
`A+B+C=[(0, -1),(3,0)]` ...(1)
Putting `lambda=-1`, we get
`A-B+C=[(4,-3),(-3,0)]`
Subtracting (2) from (1), we get
`2B[(0, -1),(3,0)]-[(4,-3),(-3,0)]=[(-4,2),(6,0)]`
`:. B=[(-2,...