The circumference of the second Bohr orbit of electron in hydrogen atom is `600nm` . The potential difference that must be applied between the plates so that the electron have the de Broglie wavelength corresponding in this circumference is

Options

A. `10^(-5) V`
B. `(5)/(3)10^(-5) V`
C. `5 xx 10^(-5) V`
D. `3 xx10^(-5) V`

Answer

D

Explanation

de Brogile wavelength of electron in hydrogen atom`= (h)/(mv) = (2 pi r_(n))/(n)`
For second Bohr orbit, `lambda = (600 xx 10^(-9))/(2) = 3000 xx 10^(-9) m`
`lambda = sqrt((150)/(V)) Å = 300 Å`
`:. V = (150)/((3000)^(2)) = (5)/(3) xx 10^(-5) V`