Which should be the value of " n " for the original excited state " A "? | TestHub

Passage / Comprehension

A sample containing a large number of atoms of unielectronic species is subjected to EMR having a wavelength of $\frac{1240}{10.2} nm$ , causing excitation in the sample. The sample contains some atoms in the ground state as well as some in one particular excited state " $A$ ". On de-excitation to the ground level, a total of 6 different wavelengths were observed, out of which 3 wavelengths were shorter than $\frac{1240}{10.2}$ . From this information, answer the questions that follow.

[Use: $h c = 1240 e V - nm$ ; I.E. of hydrogen $= 13.6 e V / a t o m$ .

Which should be the value of " $n$ " for the original excited state " $A$ "?

Answer:

2.00

Solution:

Number of spectral lines $= \frac{n ( n - 1 )}{2} = 6$

$\Rightarrow n = 4$

So atoms are excited to $n = 4$ after subjective to EMR having wavelength $= \frac{1240}{10.2 eV}$

So the possible values for original excited state $= 2, 3$

Now $3 ⟶ 4$

$3 \to 2, 2 \to 1, 3 \to 1, 4 \to 2, 4 \to 1$ have shorter wavelength than $\frac{1240}{10.2} nm$

So not possible For $4 ⟶ 2$

So $4 \to 1, 3 \to 1, 2 \to 1$ , have shorter wavelength than $\frac{1240}{10.2} nm$

$4 \to 3, 3 \to 2$ , have larger wavelength than $\frac{1240}{10.2} nm$

$4 \to 2$ , have wavelength equal to $\frac{1240}{10.2} nm$

So the value of n for original excited state $= 2$

Chemistry - Atomic Structure Question with Solution | TestHub

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