
The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to $n_x$, where x is:
A. 0
B. 1
C. -2
D. 2
Answer
575.7k+ views
Hint: The linear momentum and angular momentum of an electron are given by : $mv$ and $mvr$respectively, we have to multiply them and check the power of n from the equation. The power of n will be the value of x and the answer.
Complete step by step answer:
Angular momentum: $mvr\, = \,\dfrac{{nh}}{{2\pi }}$
Linear momentum: $mv$
Therefore, product of these two:
\[mvr\, \times mv\, = \,{m^2}{v^2}r\]
And $v\, = \,2.18\,\, \times \,{10^6}\,\dfrac{z}{n}$
$r\, = \,0.53\,\dfrac{{{n^2}}}{z}$
Therefore, putting these values in the product we check the powers of n:
Product is proportional to $\dfrac{1}{{{n^2}}} \times {n^2}\, = 1$
Hence, the product is independent of n, i.e. $n_0$
So, the correct answer is “Option A”.
Additional Information: Angular momentum of an electron by Bohr is given by $mvr$ or $\dfrac{nh}{2\pi}$ (where v is the velocity, n is the orbit in which electron is, m is mass of the electron, and r is the radius of the nth orbit). According to Bohr’s atomic model, electrons move only in those orbits where angular momentum of an electron is an integral multiple of h/2. The energy level occupied by an electron usually is called the ground state. It may move to a level of higher energy or a less stable shell by absorbing energy. This higher energy but less stable shell or level is termed as the excited state of the electron. where m is the mass, v is the velocity, r is the radius of the orbit, h is Planck's constant and n is a positive integer.
Note:
The students should apply the correct formula of the velocity and radius. The students should not waste their time in calculating the values in the product as only the dependency of n is asked in the question. Therefore if the value of n is found then the rest of the calculation is not necessary.
Complete step by step answer:
Angular momentum: $mvr\, = \,\dfrac{{nh}}{{2\pi }}$
Linear momentum: $mv$
Therefore, product of these two:
\[mvr\, \times mv\, = \,{m^2}{v^2}r\]
And $v\, = \,2.18\,\, \times \,{10^6}\,\dfrac{z}{n}$
$r\, = \,0.53\,\dfrac{{{n^2}}}{z}$
Therefore, putting these values in the product we check the powers of n:
Product is proportional to $\dfrac{1}{{{n^2}}} \times {n^2}\, = 1$
Hence, the product is independent of n, i.e. $n_0$
So, the correct answer is “Option A”.
Additional Information: Angular momentum of an electron by Bohr is given by $mvr$ or $\dfrac{nh}{2\pi}$ (where v is the velocity, n is the orbit in which electron is, m is mass of the electron, and r is the radius of the nth orbit). According to Bohr’s atomic model, electrons move only in those orbits where angular momentum of an electron is an integral multiple of h/2. The energy level occupied by an electron usually is called the ground state. It may move to a level of higher energy or a less stable shell by absorbing energy. This higher energy but less stable shell or level is termed as the excited state of the electron. where m is the mass, v is the velocity, r is the radius of the orbit, h is Planck's constant and n is a positive integer.
Note:
The students should apply the correct formula of the velocity and radius. The students should not waste their time in calculating the values in the product as only the dependency of n is asked in the question. Therefore if the value of n is found then the rest of the calculation is not necessary.
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