
If ${{v}_{n}}$ and ${{v}_{p}}$ are the velocities in nth and pth orbits respectively, then the ratio of ${{v}_{n}}:{{v}_{p}}$ is
$\begin{align}
& A.\dfrac{{{p}^{2}}}{{{n}^{2}}} \\
& B.\dfrac{{{n}^{2}}}{{{p}^{2}}} \\
& C.\dfrac{p}{n} \\
& D.\dfrac{n}{p} \\
\end{align}$
Answer
576k+ views
Hint: To solve this question we need to use the derived formula of angular momentum from de Broglie equation. According to this equation the angular momentum of the electron relative to the nucleus is quantized.
Complete answer:
In order to solve this question first we need to understand the Bohr model. This model was given by Niels Bohr and Ernest Rutherford. According to this model electrons revolve around the nucleus, consisting of protons and neutrons in circular orbits known as energy shells and can only orbit stably in certain fixed circular orbits at a discrete set of distances from the nucleus.
Let us see the formula of Angular momentum from de Broglie Equation. From this, we can obtain a formula which is used in the calculation of velocity of electrons.
$mvr=\dfrac{nh}{2\pi }$, n=number of orbits, m=Mass of electron, v=velocity of electron, r=radius of the orbit, h=Planck's constant
$\begin{align}
& v=(\dfrac{nh}{2\pi m})\dfrac{1}{r}=(\dfrac{nh}{2\pi m})(\dfrac{1}{(\dfrac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi m{{e}^{2}}})}) \\
& =(\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}h})(\dfrac{1}{n}) \\
\end{align}$
Hence, $v\propto \dfrac{1}{n}$
$\Rightarrow \dfrac{{{v}_{n}}}{{{v}_{p}}}=(\dfrac{p}{n})$
So, the correct answer is “Option C”.
Note:
Mass used in the above formula is not the molecular weight of hydrogen; it is the mass of protons of hydrogen atom. And r is the radius of the orbit. So, don’t confuse among them. The Bohr magneton is a mathematical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum. The idea of elementary magnets is due to Walter Ritz and Pierre Weiss. The Bohr magneton is the magnitude of the magnetic dipole moment of an electron orbiting an atom with such angular momentum.
Complete answer:
In order to solve this question first we need to understand the Bohr model. This model was given by Niels Bohr and Ernest Rutherford. According to this model electrons revolve around the nucleus, consisting of protons and neutrons in circular orbits known as energy shells and can only orbit stably in certain fixed circular orbits at a discrete set of distances from the nucleus.
Let us see the formula of Angular momentum from de Broglie Equation. From this, we can obtain a formula which is used in the calculation of velocity of electrons.
$mvr=\dfrac{nh}{2\pi }$, n=number of orbits, m=Mass of electron, v=velocity of electron, r=radius of the orbit, h=Planck's constant
$\begin{align}
& v=(\dfrac{nh}{2\pi m})\dfrac{1}{r}=(\dfrac{nh}{2\pi m})(\dfrac{1}{(\dfrac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi m{{e}^{2}}})}) \\
& =(\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}h})(\dfrac{1}{n}) \\
\end{align}$
Hence, $v\propto \dfrac{1}{n}$
$\Rightarrow \dfrac{{{v}_{n}}}{{{v}_{p}}}=(\dfrac{p}{n})$
So, the correct answer is “Option C”.
Note:
Mass used in the above formula is not the molecular weight of hydrogen; it is the mass of protons of hydrogen atom. And r is the radius of the orbit. So, don’t confuse among them. The Bohr magneton is a mathematical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum. The idea of elementary magnets is due to Walter Ritz and Pierre Weiss. The Bohr magneton is the magnitude of the magnetic dipole moment of an electron orbiting an atom with such angular momentum.
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