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.

Options

A. `1.675xx10^(-5)V`
B. `3.2xx10^(-5)V`
C. `1.675xx10^(-4)V`
D. `1.675xx10^(-6)V`

Answer

A

Explanation

`n=("Circmference")/("Wavelength")`
`nlamda=2pir`
`2lamda=600rArrlamda=300nm`
Let stopping potential is `V_(0)`
`eV_(0)=1//2mv^(2)`
`eV_(0)=1/2m(h/(lamdam))^(2)[lamda=h/(mv)]rArrV_(0)=h^(2)/(2mlamda^(2)e)`
`V_(0)=((6.626xx10^(-34))^(2))/(2xx(9.1xx10^(-31))xx(300xx10^(-9))^(2)xx1.6xx10^(-19))`
`V_(0)=1.675xx10^(-5)V`.