Question

What would be the atomic number of element 'X' so that the $4^{th}$ orbit around it would fit inside the 1st Bohr orbit of H-atom ?

• A. 3
• B. 4
• C. 16
• D. 25

Answer 1

Answer: (C)

16

Answer 2

The problem asks for the atomic number of an element 'X' such that its 4th Bohr orbit fits inside the 1st Bohr orbit of a Hydrogen atom.

1. Bohr Radius Formula: The radius of the $n^{th}$ Bohr orbit for a hydrogen-like atom with atomic number $Z$ is given by: $r_n(Z) = \frac{n^2 a_0}{Z}$ where $a_0$ is the Bohr radius ($0.529 \text{ Å}$).

2. Radius of 1st Bohr orbit of H-atom: For Hydrogen (H-atom), $Z=1$ and we are considering the $1^{st}$ orbit, so $n=1$. $r_1(H) = \frac{1^2 a_0}{1} = a_0$

3. Radius of 4th Bohr orbit of element 'X': For element 'X', we are considering the $4^{th}$ orbit, so $n=4$. Let its atomic number be $Z_X$. $r_4(X) = \frac{4^2 a_0}{Z_X} = \frac{16 a_0}{Z_X}$

4. Condition for fitting inside: The condition "the $4^{th}$ orbit around it would fit inside the $1^{st}$ Bohr orbit of H-atom" means that the radius of the $4^{th}$ orbit of X must be less than or equal to the radius of the $1^{st}$ Bohr orbit of H. $r_4(X) \le r_1(H)$

5. Substitute the expressions and solve for $Z_X$: $\frac{16 a_0}{Z_X} \le a_0$ Since $a_0$ is a positive constant, we can divide both sides by $a_0$: $\frac{16}{Z_X} \le 1$ Since $Z_X$ (atomic number) must be a positive integer, we can multiply both sides by $Z_X$ without changing the inequality direction: $16 \le Z_X$

6. Evaluate the options: The atomic number $Z_X$ must be greater than or equal to 16. Let's check the given options: (1) 3: $3 < 16$ (Does not satisfy) (2) 4: $4 < 16$ (Does not satisfy) (3) 16: $16 \ge 16$ (Satisfies) (4) 25: $25 \ge 16$ (Satisfies)

Both 16 and 25 satisfy the condition. However, in multiple-choice questions of this type where a specific value is asked, it usually refers to the minimum value that satisfies the condition or the threshold value. The smallest integer value for $Z_X$ that satisfies $Z_X \ge 16$ is 16. If $Z_X = 16$, then $r_4(X) = \frac{16 a_0}{16} = a_0$, which is exactly equal to $r_1(H)$, thus fitting inside (at the boundary).