15. A photoelectron experiment is performed on a sample of silicon using photons with an energy of 1.43 x 105 kJ/mole. Photoelectrons are generated that have a kinetic energy of 1.34 x 10

kJ/mole. What was the ionization energy of these electrons?

9.0×10^3K/moles

Step-by-step explanation:

ionization energy of these electrons = Energy of photons - kinetic energy of the electrons

Given

Kinetic energy of electron = 1.34×10^5

Energy of photons = 1.43 x 105 kJ/mole

Required

Ionization energy of these electrons

Ionization energy of these electrons = 1.43×10^5-1.34×10^5

ionization energy of these electrons = 0.09×10^5 kJ/mol

= 9.0 × 10^3 kJ/mol

Hence the ionization energy of these electrons is 9.0×10^3K/moles

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.