The radius of first Bohr ortbit is x. The de-Broglie wavelength of electron in 3rd orbit is `npix` where n=?

Answer

6

Explanation

Here, `lambda=2pix`
Radius of 3rd orbit is `3^(2)x=9x`
It must contain 3 de-Broglie wavelengths.
`:. 3lambda'=2pi(9x)`
`lambda'=6pix :. n=6`