The forbidden energy gap in Ge is 0.72 eV. The maximum wavelength of radiation that will generate an electron hole pair is approximately equal to x angstrom. Find (x/10). Round off answer to perfect integer. (Given: h c=12400 eV mathringA )

Question

The forbidden energy gap in Ge is 0.72 eV. The maximum wavelength of radiation that will generate an electron hole pair is approximately equal to x angstrom. Find (x/10). Round off answer to perfect integer. (Given: h c=12400 eV mathringA ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

To create hole-electron pair, 0.72=(h c/λ) ⇒ λ=(12400/0.72)=17222 mathringA

Q. The forbidden energy gap in $G e$ is $0.72 e V$ . The maximum wavelength of radiation that will generate an electron hole pair is approximately equal to $x$ angstrom. Find $\frac{x}{10}$ . Round off answer to perfect integer. (Given: $h c = 12400 e V \overset{˚}{A}$ )

To create hole-electron pair,
$0.72 = \frac{h c}{λ}$
$\Rightarrow λ = \frac{12400}{0.72} = 17222 \overset{˚}{A}$

Q. The forbidden energy gap in Ge is 0.72eV. The maximum wavelength of radiation that will generate an electron hole pair is approximately equal to x angstrom. Find 10x​. Round off answer to perfect integer. (Given: hc=12400eVA˚ )

To create hole-electron pair, 0.72=λhc​ ⇒λ=0.7212400​=17222A˚

Q. The forbidden energy gap in $G e$ is $0.72 e V$ . The maximum wavelength of radiation that will generate an electron hole pair is approximately equal to $x$ angstrom. Find $\frac{x}{10}$ . Round off answer to perfect integer. (Given: $h c = 12400 e V \overset{˚}{A}$ )

To create hole-electron pair,
$0.72 = \frac{h c}{λ}$
$\Rightarrow λ = \frac{12400}{0.72} = 17222 \overset{˚}{A}$