Answer
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Hint:According to the Bohr’s model of atomic structure, the angular momentum of the electron in an orbit is quantized. We need to use the classical definition of the angular momentum and compare it with the quantized angular momentum of the electron in the hydrogen atom.
Formula used:
\[L = mvr\]
where L is the angular momentum of the particle of mass m in a circular orbit of radius r with linear speed v.
\[L = \dfrac{{nh}}{{2\pi }}\]
where the angular momentum L is the integral multiple of the minimum angular momentum of the electron in the hydrogen atom.
Complete step by step solution:
Let the mass of the electron is m and move with linear speed v in a circular orbit of nth energy level of radius r. Then the angular momentum of the electron will be,
\[L = mvr\]
On comparing with the angular momentum as per the Bohr’s model,
\[mvr = \dfrac{{nh}}{{2\pi }}\]
If the angular speed is \[\omega \] then the angular speed is related to the linear speed as \[\omega r = v\]
So, the expression becomes,
\[m\left( {\omega r} \right)r = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow \omega = \dfrac{{nh}}{{2m\pi {r^2}}}\]
For the maximum value of the angular speed the electron should be in ground state \[n = 1\] because the angular speed is inversely proportional to the principal quantum number. The radius of the ground state orbit is \[0.53 \times {10^{ - 10}}m\]
Putting the values, we get the minimum angular speed of the electron in hydrogen atom is,
\[{\omega _{\min }} = \dfrac{{1 \times 6.626 \times {{10}^{ - 34}}}}{{2 \times 9.1 \times {{10}^{ - 34}} \times 3.14 \times {{\left( {0.53 \times {{10}^{ - 10}}} \right)}^2}}}rad/s \\ \]
\[\therefore {\omega _{\min }} = 4.13 \times {10^{16}}\,rad/s\]
Hence, the minimum angular speed of the electron in a hydrogen atom is \[4.13 \times {10^{16}}\,rad/s\].
Therefore, the correct answer is \[4.13 \times {10^{17}}\,rad/s\].
Note: The angular momentum is proportional to the principal quantum number and the angular momentum of the particle is proportional to the angular speed. So for the minimum angular speed the angular momentum will also be minimum.
Formula used:
\[L = mvr\]
where L is the angular momentum of the particle of mass m in a circular orbit of radius r with linear speed v.
\[L = \dfrac{{nh}}{{2\pi }}\]
where the angular momentum L is the integral multiple of the minimum angular momentum of the electron in the hydrogen atom.
Complete step by step solution:
Let the mass of the electron is m and move with linear speed v in a circular orbit of nth energy level of radius r. Then the angular momentum of the electron will be,
\[L = mvr\]
On comparing with the angular momentum as per the Bohr’s model,
\[mvr = \dfrac{{nh}}{{2\pi }}\]
If the angular speed is \[\omega \] then the angular speed is related to the linear speed as \[\omega r = v\]
So, the expression becomes,
\[m\left( {\omega r} \right)r = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow \omega = \dfrac{{nh}}{{2m\pi {r^2}}}\]
For the maximum value of the angular speed the electron should be in ground state \[n = 1\] because the angular speed is inversely proportional to the principal quantum number. The radius of the ground state orbit is \[0.53 \times {10^{ - 10}}m\]
Putting the values, we get the minimum angular speed of the electron in hydrogen atom is,
\[{\omega _{\min }} = \dfrac{{1 \times 6.626 \times {{10}^{ - 34}}}}{{2 \times 9.1 \times {{10}^{ - 34}} \times 3.14 \times {{\left( {0.53 \times {{10}^{ - 10}}} \right)}^2}}}rad/s \\ \]
\[\therefore {\omega _{\min }} = 4.13 \times {10^{16}}\,rad/s\]
Hence, the minimum angular speed of the electron in a hydrogen atom is \[4.13 \times {10^{16}}\,rad/s\].
Therefore, the correct answer is \[4.13 \times {10^{17}}\,rad/s\].
Note: The angular momentum is proportional to the principal quantum number and the angular momentum of the particle is proportional to the angular speed. So for the minimum angular speed the angular momentum will also be minimum.
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