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

`n lamda=2pir`
`implies2xxlamda=600`
`lamda=300nm`
Let stopping potential is `V_(0)`
`eV_(0)=1/2 mv^(2)`…………1
`lamda=h/(mv)`
`:.V=h/(mlamda)`………..2
From equation 1 and 2
`eV_(0)=1/2m(h/(lamdam))^(2)`
`V_(0)=(h^(2))/(2mlamda^(2)e)`
`=((6.626xx10^(-34))^(2))/(2xx(9.1xx1xx0^(-31))xx(300xx10^(-9))^(2)xx1.6xx10^(-19)`
`=1.675xx10^(-5)V`