The circumference of the second Bohr orbit of an electron in a hydrogen atom is `600 nm`. Calculate the potential difference to which the electron has to be accelerated to get de Broglie wavelength curresponding to this circumference.

Explanation

Number of waves `'n'=("Circumference")/("Wavelength")`
`n lamda=2pir`
`2lamda=600`
`lamda=300nm`
Let stopping potential is `V_(0)`,
`eV_(0)=(1)/(2)mv^(2)` . . .(i)
`lamda=(h)/(mv)`
`v=(h)/(lamdam)` . . . (ii)
From equations (i) and (ii),
`eV_(0)=(1)/(2)m((h)/(lamdam))^(2)`
`V_(0)=(h^(2))/(2mlamda_(2)^(2))`
`=((6.626xx10^(-34))^(2))/(2xx(9.1xx10^(-31))xx(300xx10^(-9))xx1.6xx10^(-19))`
`=1.675xx10^(-5)V`.