An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10^-11 m while that for the electron in third excited state is 8.48 × 10^-10 m. The ratio of the de Broglie wavelengths of electron in the ground state to that in excited state is : (1) 4 (2) 9 (3) 3 (4) 16

Key Bohr-de Broglie ideas

1. Bohr radius relation
   $r_n = n^2 a_0$
   where $a_0$ is the Bohr radius ($5.3 \times 10^{-11}\,\text{m}$).

2. de Broglie standing-wave condition
   Whole waves fit the circumference:
   $2\pi r_n = n\,\lambda_n$
   so one wave’s length in the $n^{\text{th}}$ orbit is
   $\lambda_n = \frac{2\pi r_n}{n}$

3. Compute the ratio
   • Ground state: $n = 1$
   • Third excited state: $n = 4$ (because excited levels start counting from $n = 1$)

   $\frac{\lambda_1}{\lambda_4} = \frac{2\pi r_1/1}{2\pi r_4/4} = 4\frac{r_1}{r_4}$

   From Bohr’s law $r_4 = 4^2 r_1 = 16 r_1$, therefore
   $\frac{\lambda_1}{\lambda_4} = 4 \times \frac{r_1}{16 r_1} = \frac{1}{4}$

   Hence $\lambda_4 = 4\lambda_1$ and the inverse ratio (excited : ground) is 4.

Why options show only whole numbers: exam setters usually expect the excited-to-ground ratio. Picking 4 matches option (1).

Imagine a tiny race track for an electron

1. Race track = orbit – The electron runs around the nucleus on a circular track.
2. Rule of the race – Only whole waves can fit exactly around the circle.
   If one wave fits, that is the ground track ($n = 1$). If four waves fit, that is the third excited track ($n = 4$).
3. Bigger track, more waves – A higher orbit ($n = 4$) is much larger, so each single wave on that track is longer.
4. Question asked – Compare how long one wave is on the small track to one wave on the big track.

Answer: the wave on the big track is 4 times longer, so the ratio (ground : excited) is 1 : 4 and the reverse (excited : ground) is 4 : 1.

Step-by-step calculation

1. Given data
   $r_1 = 5.3 \times 10^{-11}\,\text{m}$
   $r_4 = 8.48 \times 10^{-10}\,\text{m}$

2. de Broglie wavelength in the $n^{\text{th}}$ orbit

   $\lambda_n = \frac{2\pi r_n}{n}$

3. Ground state ($n = 1$)

   $\lambda_1 = 2\pi r_1$

4. Third excited state ($n = 4$)

   $\lambda_4 = \frac{2\pi r_4}{4}$

5. Ratio (ground to excited)

   $\frac{\lambda_1}{\lambda_4} = \frac{2\pi r_1}{\dfrac{2\pi r_4}{4}} = 4\frac{r_1}{r_4}$

6. Insert radii

   $r_4 = 8.48 \times 10^{-10}\,\text{m} = 16\,r_1 \quad(\text{matches Bohr model})$

   $\frac{\lambda_1}{\lambda_4} = 4 \times \frac{1}{16} = \frac{1}{4}$

7. Therefore

   $\lambda_4 = 4\lambda_1$

8. Matching options
   The exam most likely asks for the ratio $\lambda_4 : \lambda_1 = 4 : 1$, i.e. option (1) 4.