An electron is orbiting in `4^"th"` orbit in a hydrogen atom. The de-broglie's wavelangth accociated with this electron is nearly

Options

A. `1.32^@`
B. `13.2^@`
C. `15.6^@`
D. `20^@`

Answer

B

Explanation

`=p=sqrt(2nKM)`
`=sqrt(2 xx 9.1 xx 10^(-31) xx 0.85 xx 1.6 xx 10^(-19))`
`p=5 xx 10^(-25)`
`lambda=h/p=(6.63 xx 10^(-34))/(5 xx 10^(-25))=13.2 A`