Question

# The difference between $$\left ( n+2 \right )^{th}$$ Bohr radius and $$n^{th}$$ Bohr radius is equal to the $$\left ( n-2 \right )^{th}$$ Bhor radius. The value of $$n$$ is ?

Solution

## Radius of $$nth$$ Bohr orbit  is given by$$r_n = (\dfrac{h^2 \epsilon_o}{\pi m e^2})n^2$$Hence,$$r_{n}\propto n^{2}$$Thus we get $$r_{n+2}= k\left (n+2 \right )^{2}$$ and $$r_{n}=kn^{2}$$and  $$r_{n-2}= k\left (n-2 \right )^{2}$$$$\therefore \ \ \left ( n+2 \right )^{2}-n^{2}=\left ( n-2 \right )^{2}$$ $$\Rightarrow n=8$$Physics

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