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 quantisation condition
`mvr=(nh)/(2pi)` or `(h)/(mv)=(2pir)/(n)` . . . (i)
de Broglie wavelength of an electron,
`lamda=(h)/(mv)` . . . .(ii)
Equating (i) and (ii), we get `lamda=(2pir)/(n)`
For the first orbit, `r=0.53Å,n=1`
`:.lamda=(2xx3.14xx0.53)/(1)Å=3.33Å`