The difference between ${(n + 2)}^{t h}$ Bohr radius and $n^{t h}$ Bohr radius is equal to the ${(n - 2)}^{t h}$ Bhor radius. The value of $n$ is ?

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Solution

Radius of $n t h$ Bohr orbit is given by
$r_{n} = (\frac{h^{2} ϵ_{o}}{π m e^{2}}) n^{2}$
Hence,
$r_{n} \propto n^{2}$
Thus we get $r_{n + 2} = k {(n + 2)}^{2}$
and $r_{n} = k n^{2}$
and $r_{n - 2} = k {(n - 2)}^{2}$
$∴ {(n + 2)}^{2} - n^{2} = {(n - 2)}^{2}$
$\Rightarrow n = 8$

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The difference between (n+2)th Bohr radius and nth Bohr radius is equal to the (n−2)th Bhor radius. The value of n is ?

Radius of nth Bohr orbit is given byrn=(h2ϵoπme2)n2Hence,rn∝n2Thus we get rn+2=k(n+2)2 and rn=kn2and rn−2=k(n−2)2∴ (n+2)2−n2=(n−2)2 ⇒n=8

The difference between ${(n + 2)}^{t h}$ Bohr radius and $n^{t h}$ Bohr radius is equal to the ${(n - 2)}^{t h}$ Bhor radius. The value of $n$ is ?

Radius of $n t h$ Bohr orbit is given by
$r_{n} = (\frac{h^{2} ϵ_{o}}{π m e^{2}}) n^{2}$
Hence,
$r_{n} \propto n^{2}$
Thus we get $r_{n + 2} = k {(n + 2)}^{2}$
and $r_{n} = k n^{2}$
and $r_{n - 2} = k {(n - 2)}^{2}$
$∴ {(n + 2)}^{2} - n^{2} = {(n - 2)}^{2}$
$\Rightarrow n = 8$