The radius of the 2^ nd orbit of He^+ of Bohr's model is r1 and t… — Solution | JEE Physics

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The radius of the $2^{\text{nd}}$ orbit of $He^+$ of Bohr's model is $r_1$ and that of the fourth orbit of $Be^{3+}$ is represented as $r_2$ . Now the ratio $\frac{r_2}{r_1}$ is $x : 1$ . The value of $x$ _____

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: The radius of the $2^{\text{nd}}$ orbit of $He^+$ is $r_1$ and the radius of the fourth orbit of $Be^{3+}$ is $r_2$ .

Find: The value of $x$ in $\frac{r_2}{r_1} = x$ .

In Bohr's model, the radius of an orbit is proportional to

\frac{n^2}{Z}

where $n$ is the orbit number and $Z$ is the atomic number.

So,

\frac{r_2}{r_1} = \left( \frac{n_2}{n_1} \right)^2 \times \frac{Z_1}{Z_2}

Substituting the given values,

n_1 = 2, \quad n_2 = 4, \quad Z_1 = 2, \quad Z_2 = 4

Therefore,

\frac{r_2}{r_1} = \left( \frac{4}{2} \right)^2 \times \frac{2}{4}

\frac{r_2}{r_1} = 2^2 \times \frac{1}{2} = 4 \times \frac{1}{2} = 2

Therefore, the value of $x$ is $2$ .

Direct Ratio Approach

Given: $r \propto \frac{n^2}{Z}$ .

Find: $x = \frac{r_2}{r_1}$ .

Using proportionality directly,

r_1 \propto \frac{2^2}{2}, \qquad r_2 \propto \frac{4^2}{4}

So,

\frac{r_2}{r_1} = \frac{\frac{16}{4}}{\frac{4}{2}} = \frac{4}{2} = 2

This works because for hydrogen-like species in Bohr's model, orbit radius depends only on $n^2$ and inversely on $Z$ .

Therefore, the value of $x$ is $2$ .

Common mistakes

Using radius proportional to $\frac{n}{Z}$ instead of $\frac{n^2}{Z}$ . This gives an incorrect dependence on orbit number. Always use Bohr's radius relation with the square of $n$ .
Taking the atomic number of $Be^{3+}$ as $3$ because of the charge. This is wrong because $Z$ is the number of protons, so for beryllium it is $4$ . Use charge only to identify that the species is hydrogen-like.
Reversing $\frac{r_2}{r_1}$ as $\frac{r_1}{r_2}$ during substitution. This changes the final numerical value. Write clearly which orbit corresponds to $r_1$ and which corresponds to $r_2$ before forming the ratio.

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