JEE Chemistry 2025 Question with Solution

Given: Bohr's model of hydrogen atom.

Find: Which statement about orbit radius is incorrect.

For Bohr's model of hydrogen atom, radius of orbit is proportional to the square of the principal quantum number:

$$r \propto n^2$$

So, for comparing two orbits,

$$\frac{r_{n_2}}{r_{n_1}} = \frac{n_2^2}{n_1^2}$$

Now check each statement:

$$\frac{r_3}{r_1} = \left(\frac{3}{1}\right)^2 = 9$$

So option A is correct.

$$\frac{r_8}{r_4} = \left(\frac{8}{4}\right)^2 = 4$$

So option B is correct.

$$\frac{r_6}{r_4} = \left(\frac{6}{4}\right)^2 = \frac{36}{16} = \frac{9}{4} = 2.25$$

So the radius of the $6$th orbit is not three times that of the $4$th orbit. Hence option C is incorrect.

$$\frac{r_4}{r_2} = \left(\frac{4}{2}\right)^2 = 4$$

So option D is correct.

Therefore, the incorrect statement is option C.

Given: For hydrogen atom, Bohr orbit radius depends on principal quantum number $n$.

Find: The incorrect statement among the four options.

Concept used:

$$r_n = a_0 n^2$$

Thus,

$$r_n \propto n^2$$

where $a_0$ is the Bohr radius.

To compare radii of two orbits,

$$\frac{r_{n_2}}{r_{n_1}} = \frac{n_2^2}{n_1^2}$$

Step 1: For the $3$rd and $1$st orbits,

$$\frac{r_3}{r_1} = \frac{3^2}{1^2} = \frac{9}{1} = 9$$

Hence the $3$rd orbit radius is nine times the $1$st orbit radius. This statement is correct.

Step 2: For the $8$th and $4$th orbits,

$$\frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4$$

Hence the $8$th orbit radius is four times the $4$th orbit radius. This statement is correct.

Step 3: For the $6$th and $4$th orbits,

$$\frac{r_6}{r_4} = \frac{6^2}{4^2} = \frac{36}{16} = \frac{9}{4} = 2.25$$

Hence the $6$th orbit radius is $2.25$ times the $4$th orbit radius, not three times. This statement is incorrect.

Step 4: For the $4$th and $2$nd orbits,

$$\frac{r_4}{r_2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$$

Hence the $4$th orbit radius is four times the $2$nd orbit radius. This statement is correct.

Therefore, the only incorrect statement is C.