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

Options

A. 1.67 Å
B. 3.33 Å
C. 1.06 Å
D. 1.53 Å

Answer

B

Explanation

According to Bohr.s quantisation condition
`mvr=(nh)/(2pi)orh/(m upsilon)=(2 pi r)/n" "...(i)`
De-Brogle wavelangth of an electron
`lambda=h/(m upsilon) ...(ii)`
Equating (i) and (ii), we get
`lambda=(2 pi r)/n`
`r=0.53 dotA, n=1`
`:. lambda=(2 xx 3.14 xx 0.53)/1 dotA=3.33 dotA`