MHT CET · Chemistry · Structure of Atom
What is the energy of an electron in stationary state corresponding to \(\mathrm{n}=2\) ?
- A \(-1.45 \times 10^{-18} \mathrm{~J}\)
- B \(-0.545 \times 10^{-18} \mathrm{~J}\)
- C \(-3.45 \times 10^{-18} \mathrm{~J}\)
- D \(-2.5 \times 10^{-18} \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(-0.545 \times 10^{-18} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The energy of the stationary state is given by
\(
\begin{aligned}
& \mathrm{E}=-\mathrm{R}_{\mathrm{H}} \frac{1}{\mathrm{n}^2} \\
& \mathrm{R}_{\mathrm{H}}=\text { Rydberg constant }=2.18 \times 10^{-18} \mathrm{~J} \\
& \mathrm{E}=-2.18 \times 10^{-18} \times \frac{1}{2^2} \\
& =-0.545 \times 10^{-18} \mathrm{~J} .
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{E}=-\mathrm{R}_{\mathrm{H}} \frac{1}{\mathrm{n}^2} \\
& \mathrm{R}_{\mathrm{H}}=\text { Rydberg constant }=2.18 \times 10^{-18} \mathrm{~J} \\
& \mathrm{E}=-2.18 \times 10^{-18} \times \frac{1}{2^2} \\
& =-0.545 \times 10^{-18} \mathrm{~J} .
\end{aligned}
\)
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