Question

An atomic emission spectrum of hydrogen shows three wavelengths:

121.5 nm,102.6 nm, and 97.23 nm

Assign these wavelengths to transitions in the hydrogen atom.

Drag the appropriate items to their respective bins.

Answer 1

We have that wavelengths to transitions in the hydrogen atom.

$nf=2,ni=1 \rightarrow 121.5nm\\\\ni=1,nf=3 \rightarrow 102.6nm\\\\\ni=1,nf=4 \rightarrow 97.2nm$

From the question we are told that

121.5 nm,102.6 nm, and 97.23 nm

Generally the equation for wavelength  is mathematically given as

$\frac{1}{\lambda}=R_H*(\frac{1}{nf^2}-\frac{1}{nf^2})\\\\\frac{1}{\lambda}=1.097*10^7*\frac{1}{nf^2}-\frac{1}{ni^2}$

Therefore

For nf=2,ni=1

$\frac{1}{\lambda}=1.097*10^7*\frac{1}{2^2}-\frac{1}{1^2}\\\\\frac{1}{\lambda}=121.5nm$

For ni=1,nf=3

$\frac{1}{\lambda}=1.097*10^7*\frac{3}{1^2}-\frac{1}{1^2}\\\\\frac{1}{\lambda}=102.6nm$

For ni=1,nf=4

$\frac{1}{\lambda}=1.097*10^7*\frac{1}{4^2}-\frac{1}{1^2}\\\\\frac{1}{\lambda}=97.2nm$

Therefore

The Correct slots are

$nf=2,ni=1 \rightarrow 121.5nm\\\\ ni=1 ,nf = 3  \rightarrow 102.6nm\\\\\ni=1,nf=4   \rightarrow  97.2 nm$

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