Question

Calculate the long wavelength limit of extrinsic semiconductor if ionization energy is 0.02 eV

Answer 1

E = hc/wavelength

If energy is given in terms of eV, the formula becomes

lambda = 12400/E

12400/0.02

lambda = 620,000 angstrom or

lambda = 6.2 x 10^ -4 m

Answer 2

994.6×$10^{-26}$. is  long wavelength limit of extrinsic semiconductor if ionization energy is 0.02 eV.

Given,

E = 0.02eV

To find,

Wavelength = ?

Solution,

Ionization energy,

• Minimum energy which electron needs in gas ion or atom form and absorbs to be out of the nucleus influence.
• This process is majorly termed as Ionization energy and is also Endothermic process.

Formula used,

$E = \frac{hc}{wavelength}$

λ = $\frac{hc}{E}$

λ = $\frac{6.63[10^{-34}] 3[10^{8}]}{0.02} = 994.6[10^{-26}]$

λ = 994.6×$10^{-26}$.

Therefore, 994.6×$10^{-26}$. is  long wavelength limit of extrinsic semiconductor if ionization energy is 0.02 eV.