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GUJCET · Physics · Atoms
A difference of 5.4 eV separates two energy levels in atom.What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
(Take, \(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}, h=6.625 \times 10^{-34} \mathrm{Js}\) )
- A \(5.6 \times 10^{14} \mathrm{~Hz}\)
- B \(1.304 \times 10^{15} \mathrm{~Hz}\)
- C \(5.6 \times 10^{15} \mathrm{~Hz}\)
- D \(1.304 \times 10^{14} \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(B) \(1.304 \times 10^{15} \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
(B) \(1.304 \times 10^{15} \mathrm{~Hz}\)
\(\mathrm{E}_{2}-\mathrm{E}_{1}=h v\)
\(
\begin{array}{ll}
\therefore & v=\frac{5.4 \times 1.6 \times 10^{-19}}{6.625 \times 10^{-34}} \\
\therefore & v=1.304 \times 10^{15} \mathrm{~Hz}
\end{array}
\)
\(\mathrm{E}_{2}-\mathrm{E}_{1}=h v\)
\(
\begin{array}{ll}
\therefore & v=\frac{5.4 \times 1.6 \times 10^{-19}}{6.625 \times 10^{-34}} \\
\therefore & v=1.304 \times 10^{15} \mathrm{~Hz}
\end{array}
\)
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