
In a hydrogen atom following Bohr's postulates, the product of linear momentum and angular momentum is proportional to \[{n^x}\] where n is the orbit number. Then x is
A. 0
B. 2
C. -2
D. 1
Answer
162.9k+ views
Hint: In order to solve this problem, we need to find the linear momentum and the angular momentum of the electron in the hydrogen atom as per Bohr’s postulates.
Formula used:
\[{v_n} = \left( {2.18 \times {{10}^6}m/s} \right)\left( {\dfrac{Z}{n}} \right)\]
where \[{v_n}\] is the speed of an electron in the nth orbit of an atom with atomic number Z.
\[L = \dfrac{{nh}}{{2\pi }}\]
where L is the angular momentum of an electron in the nth orbit of an atom.
Complete step by step solution:
According to Bohr’s model of atoms, the electron revolves around the nucleus of the atom with uniform speed. In a hydrogen atom, there is only one electron revolving around the nucleus under the influence of attractive force applied by the positive nucleus of the hydrogen.
The speed of the electron in nth orbit of atom is given as,
\[{v_n} = \left( {2.18 \times {{10}^6}\dfrac{Z}{n}} \right)m/s\]
Here, Z is the atomic number of the atom.
For hydrogen atoms, the atomic number is 1.
So, the speed of electron in nth orbit of hydrogen atom will be,
\[{v_n} = \dfrac{{\left( {2.18 \times {{10}^6}m/s} \right)}}{n}\]
So, the linear momentum of the electron will be,
\[P = m{v_n}\]
\[\Rightarrow P = m\left( {\dfrac{{2.18 \times {{10}^6}}}{n}} \right) \\ \]
The angular momentum of the electron as per Bohr’s postulates is the integral multiple of \[\dfrac{h}{{2\pi }}\]
For the electron in hydrogen atom in nth orbit, the angular momentum will be,
\[L = \dfrac{{nh}}{{2\pi }}\]
Then the product of the linear momentum of the electron in hydrogen atom and the angular momentum of the electron in the same energy level of hydrogen atom is,
\[LP = \left( {\dfrac{{nh}}{{2\pi }}} \right)\left( {\dfrac{{2.18 \times {{10}^6}m}}{n}} \right) \\ \]
\[\Rightarrow LP = {n^0}\left( {\dfrac{{2.18 \times {{10}^6}mh}}{{2\pi }}} \right) \\ \]
So, the product of the linear momentum of the electron in hydrogen atom and the angular momentum is proportional to \[{n^0}\], but in the question is given that the product of the linear momentum and the angular momentum of the electron in the hydrogen atom is \[{n^x}\]. On comparing both the terms, we get \[x = 0\]. Hence, the value of x is zero.
Therefore,the correct option is A.
Note: The angular momentum is always perpendicular to the linear momentum vector. Bohr's postulate is valid for hydrogen or hydrogen line atoms. The hydrogen- like atom is an atom with a single electron orbiting the nucleus.
Formula used:
\[{v_n} = \left( {2.18 \times {{10}^6}m/s} \right)\left( {\dfrac{Z}{n}} \right)\]
where \[{v_n}\] is the speed of an electron in the nth orbit of an atom with atomic number Z.
\[L = \dfrac{{nh}}{{2\pi }}\]
where L is the angular momentum of an electron in the nth orbit of an atom.
Complete step by step solution:
According to Bohr’s model of atoms, the electron revolves around the nucleus of the atom with uniform speed. In a hydrogen atom, there is only one electron revolving around the nucleus under the influence of attractive force applied by the positive nucleus of the hydrogen.
The speed of the electron in nth orbit of atom is given as,
\[{v_n} = \left( {2.18 \times {{10}^6}\dfrac{Z}{n}} \right)m/s\]
Here, Z is the atomic number of the atom.
For hydrogen atoms, the atomic number is 1.
So, the speed of electron in nth orbit of hydrogen atom will be,
\[{v_n} = \dfrac{{\left( {2.18 \times {{10}^6}m/s} \right)}}{n}\]
So, the linear momentum of the electron will be,
\[P = m{v_n}\]
\[\Rightarrow P = m\left( {\dfrac{{2.18 \times {{10}^6}}}{n}} \right) \\ \]
The angular momentum of the electron as per Bohr’s postulates is the integral multiple of \[\dfrac{h}{{2\pi }}\]
For the electron in hydrogen atom in nth orbit, the angular momentum will be,
\[L = \dfrac{{nh}}{{2\pi }}\]
Then the product of the linear momentum of the electron in hydrogen atom and the angular momentum of the electron in the same energy level of hydrogen atom is,
\[LP = \left( {\dfrac{{nh}}{{2\pi }}} \right)\left( {\dfrac{{2.18 \times {{10}^6}m}}{n}} \right) \\ \]
\[\Rightarrow LP = {n^0}\left( {\dfrac{{2.18 \times {{10}^6}mh}}{{2\pi }}} \right) \\ \]
So, the product of the linear momentum of the electron in hydrogen atom and the angular momentum is proportional to \[{n^0}\], but in the question is given that the product of the linear momentum and the angular momentum of the electron in the hydrogen atom is \[{n^x}\]. On comparing both the terms, we get \[x = 0\]. Hence, the value of x is zero.
Therefore,the correct option is A.
Note: The angular momentum is always perpendicular to the linear momentum vector. Bohr's postulate is valid for hydrogen or hydrogen line atoms. The hydrogen- like atom is an atom with a single electron orbiting the nucleus.
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