Question
Explain the quantisation of angular momentum for an electron orbiting a nucleus.
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Write second postulate of Bohr's model and De-Broglie's explanation of it.
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Write second postulate of Bohr's model and De-Broglie's explanation of it.
✓
Answer
• Bohr's second postulate :
→An electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2 \pi}$.
Where, $h$ is Planck's constant
$h=6.625 \times 10^{-34} JS$
→Thus the angular momentum of the electron
$L =\frac{n h}{2 \pi} \text { Where, } n=1,2,3 \ldots \ldots$
• De - Broglie's explanation :
→According to de-Broglie's hypothesis even matter particles like electrons have wave nature. Its practical explanation was given by Davisson and Germer, from which de-Broglie argued that the electron in its circular orbit must be seen as a particle wave.
→When the tensioned wire is plucked tied to a rigid support on both ends, a vast number of wavelengths are excited. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string. It means standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength or any integral number of wavelength.
→Waves with other wavelengths interfere with themselves upon reflection and their amplitudes rapidly drop to zero.
→For an electron moving in $n^{\text {th }}$ circular orbit of radius $r_n$, the total distance is the circumference of the orbit. Thus,
$2 \pi r_n=n \lambda$
Where $n=1,2,3, \ldots \ldots .$.
→But the de-Broglie wavelength $\lambda=\frac{h}{p}$
Where $p =$ momentum of electron. If the speed of the electron is much less than the speed of light, the momentum is $= m v_{ n }$
$\therefore \lambda=\frac{h}{m v_n}$
→Form equation (1) and (2),
$\begin{array}{l}
\therefore 2 \pi r_n=\frac{n h}{m v_n} \\
\therefore m v_n r_n=\frac{n h}{2 \pi}
\end{array}$
→This is the quantum condition proposed by Bohr for the angular momentum of the electron.
→Thus, de-Broglie hypothesis provided an explanation for Bohr's second postulate for the quantisation of angular momentum of the orbiting electron.
→An electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2 \pi}$.
Where, $h$ is Planck's constant
$h=6.625 \times 10^{-34} JS$
→Thus the angular momentum of the electron
$L =\frac{n h}{2 \pi} \text { Where, } n=1,2,3 \ldots \ldots$
• De - Broglie's explanation :
→According to de-Broglie's hypothesis even matter particles like electrons have wave nature. Its practical explanation was given by Davisson and Germer, from which de-Broglie argued that the electron in its circular orbit must be seen as a particle wave.
→When the tensioned wire is plucked tied to a rigid support on both ends, a vast number of wavelengths are excited. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string. It means standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength or any integral number of wavelength.
→Waves with other wavelengths interfere with themselves upon reflection and their amplitudes rapidly drop to zero.
→For an electron moving in $n^{\text {th }}$ circular orbit of radius $r_n$, the total distance is the circumference of the orbit. Thus,
$2 \pi r_n=n \lambda$
Where $n=1,2,3, \ldots \ldots .$.
→But the de-Broglie wavelength $\lambda=\frac{h}{p}$
Where $p =$ momentum of electron. If the speed of the electron is much less than the speed of light, the momentum is $= m v_{ n }$
$\therefore \lambda=\frac{h}{m v_n}$
→Form equation (1) and (2),
$\begin{array}{l}
\therefore 2 \pi r_n=\frac{n h}{m v_n} \\
\therefore m v_n r_n=\frac{n h}{2 \pi}
\end{array}$
→This is the quantum condition proposed by Bohr for the angular momentum of the electron.
→Thus, de-Broglie hypothesis provided an explanation for Bohr's second postulate for the quantisation of angular momentum of the orbiting electron.
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