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

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Modern Physics

The radius of

2^{nd}

orbit of

He^{+}

of Bohr's model is

r_{1}

and that of fourth orbit of

Be^{3 +}

is represented as

r_{2}

. Now the ratio

\frac{r _{2}}{r _{1}}

x : 1

. The value of

x

is ______.

Answer: 2

Solution

In Bohr's model, the radius of the

n^{th}

orbit is:

r_{n} \propto \frac{n ^{2}}{Z}

For

He^{+}

(Z=2), second orbit (n=2):

r_{1} \propto \frac{2 ^{2}}{2} = 2

For

Be^{3 +}

(Z=4), fourth orbit (n=4):

r_{2} \propto \frac{4 ^{2}}{4} = 4

Therefore:

\frac{r _{2}}{r _{1}} = \frac{4}{2} = 2 = \frac{x}{1}

Hence,

x = 2

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer

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