Question

In which of the following systems will be radius of the first orbit (n=1) be minimum:

• A. Doubly ionized lithium
• B. Singly ionized helium
• C. Deuterium atom
• D. Hydrogen atom

Answer 1

The Correct Option is A

The problem at hand is to determine which of the given atomic systems has the smallest radius for the first orbit (n=1). To find the solution, we must consider the formula for the radius of an electron's orbit in a hydrogen-like atom:

The radius of the n^th orbit is given by the formula:

r_n = \frac{n^2 \cdot h^2}{4 \pi^2 \cdot m_e \cdot k \cdot Z \cdot e^2}

Here:

• r_n is the radius of the n^th orbit.
• n is the principal quantum number (n=1 for the first orbit).
• h is Planck's constant.
• m_e is the mass of the electron.
• k is Coulomb's constant.
• Z is the atomic number.
• e is the charge of the electron.

The most important factor here for determining the radius of the first orbit is the atomic number Z, as it is directly inversely proportional to the radius. The greater the value of Z, the smaller the orbit will be.

Now, let's consider the atomic configurations provided in the options:

• Hydrogen atom: Z = 1
• Deuterium atom: Z = 1 (Deuterium is an isotope of hydrogen, thus Z remains the same.)
• Singly ionized helium (He⁺): Z = 2
• Doubly ionized lithium (Li²⁺): Z = 3

Among these, the doubly ionized lithium has the highest atomic number (Z = 3), meaning it will have the smallest first orbit radius.

The correct answer is, therefore, the doubly ionized lithium (Li²⁺) as it has the highest effective nuclear charge, leading to the minimum radius of the first orbit.