Question

The figure shows standing de Broglie waves due to the revolution of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their usual meanings)

• A. \(\frac{h^2\in_0}{\pi me^2}\)
• B. \(\frac{4h^2\in_0}{\pi me^2}\)
• C. \(\frac{36h^2\in^{2}_0}{\pi me^2}\)
• D. \(\frac{9h^2\in_0}{\pi me^2}\)

Answer 1

The Correct Option is D

Step 1: De Broglie Condition

Circumference = nλ

$2 \pi r = n \lambda$

Step 2: De Broglie Wavelength

$\lambda = \frac{h}{mv}$

Step 3: Substitute λ in Step 1

$2 \pi r = n \frac{h}{mv}$

Step 4: Electrostatic Force = Centripetal Force

$\frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r}$

$v^2 = \frac{e^2}{4\pi\epsilon_0 mr}$

Step 5: Express v

$v = \frac{e}{\sqrt{4\pi\epsilon_0 mr}}$

Step 6: Substitute v in Step 3

$2 \pi r = n \frac{h}{m \frac{e}{\sqrt{4\pi\epsilon_0 mr}}}$

$2 \pi r = n \frac{h \sqrt{4\pi\epsilon_0 mr}}{me}$

Step 7: Solve for r

$r^{1/2} = n \frac{h \sqrt{4\pi\epsilon_0 m}}{2\pi me}$

$r = n^2 \frac{h^2 4\pi\epsilon_0 m}{(2\pi me)^2} = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$

Step 8: Determine n from figure

Count wavelengths in circumference, $n=3$.

Step 9: Substitute n=3 in r

$r = \frac{3^2 h^2 \epsilon_0}{\pi m e^2} = \frac{9 h^2 \epsilon_0}{\pi m e^2}$

Final Answer: The final answer is $\frac{9 h^2 \epsilon_0}{\pi m e^2}$

Answer 2

The de Broglie wavelength for the electron is given by:

$\lambda = \frac{h}{mv}$

For standing waves, the circumference of the orbit must be an integer multiple of the wavelength. Therefore, the condition for the orbit radius $r$ is:

$2\pi r = n \lambda$

Substituting for the de Broglie wavelength $\lambda$, we get:

$2\pi r = n \frac{h}{mv}$

This leads to an expression for the radius in terms of fundamental constants, and for the ground state (where $n = 1$), the radius is given by:

$r = \frac{9h^2 \epsilon_0}{\pi m e^2}$

Thus, the correct expression for the orbit radius is option $r = \frac{9h^2 \epsilon_0}{\pi m e^2}$.