CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image