Chemistry, 15.10.2019 21:30 landsjr69

The emission spectrum of an element
a) is different from any other element's absorption spectrum
b) may be identical to the emission spectrum of another element
c) may be identical to the absorption spectrum of another element

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Answer from: axleon11

The emission spectrum of an element may be identical to the absorption spectrum of another element.

Option C

Explanation:

The emission spectrum acts as fingerprint of each element. As each element consists of discrete energy levels, so does the emission spectrum. As emission spectrum occurs when electron emit from higher energy orbitals to lower energy orbitals releasing an electromagnetic radiations in the form of loss of energy.

And absorption spectrum of an element is obtained when the element absorbs a specific wavelength of light incident to it. So, in absorption spectrum, the wavelengths which are absorbed by element will show a black line in its position.

Thus, emission spectrum obtained for an element is inversely related to the absorbance spectrum of the same element. If we represent it mathematically, it will look like ,

$\text { Emission spectrum of an Element }=\frac{1}{\text { Absorption spectrum of an Element }}$

So, if we consider two different elements $E_{1}$ and $E_{2}$ , then

$\begin{array}{l}{\text { Emission spectrum of } E_{1}=\frac{1}{\text { Absorption spectrum of } E_{1}}} \\ {\text { Emission spectrum of } E_{2}=\frac{1}{\text { Absorption spectrum of } E_{2}}}\end{array}$

Now if we take ratio of emission spectrums, we will get

$\frac{\text {Emission spectrum of } E_{1}}{\text {Emission spectrum of } E_{2}}=\frac{\text {Absorption spectrum of } E_{2}}{\text {Absorption spectrum of } E_{1}}$

So, according to this, we can say that emission spectrum of an element can be identical to absorption spectrum of any other element.

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$