Question:
If the time period of an electron in the ground state of hydrogen atom is \(1.5\times10^{-16}\) s, then the time period of the electron in the third excited state of hydrogen atom is
If the time period of an electron in the ground state of hydrogen atom is \(1.5\times10^{-16}\) s, then the time period of the electron in the third excited state of hydrogen atom is
Show Hint
In the Bohr model of the hydrogen atom,
\[
\boxed{
T_n\propto n^3.
}
\]
Remember that
\[
\boxed{
\text{Third excited state } \Rightarrow n=4.
}
\]
Updated On: Jul 18, 2026
- \(2.4\times10^{-15}\) s
- \(9.6\times10^{-15}\) s
- \(6.4\times10^{-15}\) s
- \(4.8\times10^{-15}\) s
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The Correct Option is B
Solution and Explanation
Step 1: Identify the orbit number.
The third excited state corresponds to
\[
n=4.
\]
Step 2: Use the relation for time period. For the hydrogen atom, \[ T_n\propto n^3. \] Hence, \[ \frac{T_4}{T_1} = \left(\frac41\right)^3 = 64. \] Therefore, \[ T_4 = 64\times1.5\times10^{-16} = 96\times10^{-16} = 9.6\times10^{-15}\text{ s}. \]
Step 3: Write the answer. Hence, \[ \boxed{9.6\times10^{-15}\text{ s}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
Step 2: Use the relation for time period. For the hydrogen atom, \[ T_n\propto n^3. \] Hence, \[ \frac{T_4}{T_1} = \left(\frac41\right)^3 = 64. \] Therefore, \[ T_4 = 64\times1.5\times10^{-16} = 96\times10^{-16} = 9.6\times10^{-15}\text{ s}. \]
Step 3: Write the answer. Hence, \[ \boxed{9.6\times10^{-15}\text{ s}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
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