JEE Physics 2023 Question with Solution

Given: Radius of the first orbit is $a_0$. We need the de Broglie wavelength of the electron in the $3^{\text{rd}}$ orbit.

Find: de Broglie wavelength in the $n=3$ orbit.

Using the relation shown in the solution:

$$\lambda = \frac{2\pi r}{n}$$

For hydrogen atom, the radius of the $n^{\text{th}}$ orbit is:

$$r_n = a_0 n^2$$

Substituting this into the wavelength expression:

$$\lambda = \frac{2\pi a_0 n^2}{n}$$ $$\lambda = 2\pi a_0 n$$

For the $3^{\text{rd}}$ orbit, $n=3$:

$$\lambda = 2\pi a_0 \times 3 = 6\pi a_0$$

Therefore, the de Broglie wavelength is $6\pi a_0$. The correct option is C.

Given: The first Bohr orbit radius is $a_0$.

Find: de Broglie wavelength of the electron in the $3^{\text{rd}}$ orbit.

From de Broglie relation:

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

For an electron in the $n^{\text{th}}$ Bohr orbit, the momentum is:

$$p = mv = \frac{nh}{2\pi r}$$

Therefore,

$$\lambda = \frac{h}{p} = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n}$$

For the third orbit,

$$r = 3^2 a_0 = 9a_0$$

Now substitute $r = 9a_0$ and $n=3$:

$$\lambda = \frac{2\pi (9a_0)}{3}$$ $$\lambda = 6\pi a_0$$

Therefore, the correct answer is $6\pi a_0$, so the correct option is C.