QJEEAdv04Ch442f: JEE Advanced 2004 Chemistry - Class 11 JEEAdv04Ch442f Solution: The radius of which of the following orbit is same as that of…

To compare the radius of these different orbits to the first Bohr's orbit of the hydrogen atom, we can use the formula for the radius of the nth orbit in the hydrogen-like atom, which is given by:

Where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number,
• $h$ is Planck's constant,
• $k$ is Coulomb's constant,
• $m$ is the mass of the electron,
• $e$ is the charge of the electron,
• $Z$ is the atomic number or the number of protons in the nucleus (which also equals the charge of the nucleus in a hydrogen-like atom).

For the first Bohr orbit of the hydrogen atom ( $n=1$ and $Z=1$ ), the radius is:

We must compare this with the options provided. The formula simplifies to identifying where the $n^2/Z$ ratio for each option is equal to 1, as it is for hydrogen's first orbit.

• Option A: He⁺ ( $Z=2$ , $n=2$ )

  $\frac{n^2}{Z} = \frac{2^2}{2} = 2$

• Option B: Li²⁺ ( $Z=3$ , $n=2$ )

  $\frac{n^2}{Z} = \frac{2^2}{3} \approx 1.33$

• Option C: Li²⁺ ( $Z=3$ , $n=3$ )

  $\frac{n^2}{Z} = \frac{3^2}{3} = 3$

• Option D: Be³⁺ ( $Z=4$ , $n=2$ )

  $\frac{n^2}{Z} = \frac{2^2}{4} = 1$

Thus, the radius is the same as that of the first Bohr's orbit of hydrogen atom only for Option D: Be³⁺ ( $n=2$ ), because in this case, the ratio $n^2/Z = 1$ , matching the ratio for hydrogen's first orbit.