Question 462: Atoms & Nuclei

Question

If '$$E$$' and '$$L$$' denote the magnitude of total energy and angular momentum of revolving electron in $$\mathrm{n}^{	ext {th }}$$ Bohr orbit, then

Accepted Answer

Let's analyze the relationship between the energy (E) and angular momentum (L) of an electron in the $$n^{	ext{th}}$$ Bohr orbit according to Bohr's theory.
The energy of an electron in the $$n^{	ext{th}}$$ Bohr orbit is given by:
$$E = -\frac{k e^2}{2r_n}$$
where $$k$$ is a constant, $$e$$ is the electronic charge, and $$r_n$$ is the radius of the $$n^{	ext{th}}$$ orbit.
The angular momentum ($$L$$) of the electron in the $$n^{	ext{th}}$$ Bohr orbit is quantized and given by:
$$L = n\hbar$$
where $$\hbar$$ is the reduced Planck's constant, and $$n$$ is the principal quantum number.
The radius of the $$n^{	ext{th}}$$ orbit ($$r_n$$) according to Bohr's model is given by:
$$r_n = n^2 \frac{\hbar^2}{k' m_e e^2}$$
where $$m_e$$ is the mass of the electron and $$k'$$ is another constant.
By substituting the expression for $$r_n$$ in the equation for energy $$E$$, we get:
$$E = -\frac{k e^2}{2 \left(n^2 \frac{\hbar^2}{k' m_e e^2}ight)}$$
$$E \propto -\frac{1}{n^2}$$
Now, since $$L = n\hbar$$, it implies that $$n \propto L$$. Thus, the relationship between energy ($$E$$) and angular momentum ($$L$$) can be rewritten in terms of $$L$$:
$$E \propto -\frac{1}{L^2}$$
Therefore, among the provided options, the correct one is:
Option C: $$\mathrm{E} \propto \mathrm{L}^{-2}$$

Answer

$$\mathrm{E} \propto \mathrm{L}^{-1}$$

Answer

$$\mathrm{E} \propto \mathrm{L}$$

Answer

$$\mathrm{E} \propto \mathrm{L}^{-2}$$

Answer

$$\mathrm{E} \propto \mathrm{L}^2$$

Atoms & Nuclei

If '$E' and 'L' denote the magnitude of total energy and angular momentum of revolving electron in \mathrm{n}^{\text {th }}$ Bohr orbit, then

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