The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is (radius of the first orbit of hydrogen atom `=0.53Å`

Options

A. `1.67Å`
B. `3.33Å`
C. `1.06Å`
D. `0.53Å`

Answer

B

Explanation

According to Bohr.s quantisationof angular momentum
`mvr=(nh)/(2pi)`
or `(h)/(mv)=(2pir)/(n) " "...(i)`
de-Broglie wavelength
`lambda=(h)/(mv) " "...(ii)`
From Eqs (i) and(ii), we get
Wavelength `lambda=(2pir)/(n)`
`=(2pxx pi xx0.53Å)/(1)`
`=3.33Å`