KCET · Physics · Atomic Physics
What is the wavelength of light for the least energetic photon emitted in the Lyman series of
the hydrogen spectrum, (take hc \( =1240 \mathrm{eV} \mathrm{mm} \) )
- A \( 82 \mathrm{~mm} \)
- B \( 102 \mathrm{~mm} \)
- C \( 122 \mathrm{~mm} \)
- D \( 150 \mathrm{~mm} \)
Answer & Solution
Correct Answer
(C) \( 122 \mathrm{~mm} \)
Step-by-step Solution
Detailed explanation
We know, \( \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)
For Lyman series, \( n_{1}=1 \) and \( n 2=2 \) is least energetic. Therefore,
\[
\begin{array}{l}
\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=1.097 \times 10^{7}\left(\frac{4-1}{4}\right) \\
=\frac{3}{4} \times 1.09 \times 10^{7}
\end{array}
\]
Thus \( \lambda=\frac{4}{3 \times 1.097 \times 10^{7}}=122 \mathrm{~nm} \)
For Lyman series, \( n_{1}=1 \) and \( n 2=2 \) is least energetic. Therefore,
\[
\begin{array}{l}
\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=1.097 \times 10^{7}\left(\frac{4-1}{4}\right) \\
=\frac{3}{4} \times 1.09 \times 10^{7}
\end{array}
\]
Thus \( \lambda=\frac{4}{3 \times 1.097 \times 10^{7}}=122 \mathrm{~nm} \)
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