Derive an expression for the total energy of electron in `n^(th)` Bohr orbit. Hence show that energy of the electron is inversely proportional to the square of principal quantum number. Also define binding energy.
The photoelectric threshold wavelength of a metal is is `230nm`. Determine the maximum kinetic energy in joule and in `eV` of the ejected electron for the metal surface when it is exposed to a radiation of wavelength `180nm`.
[Plank's constant: `h=6.63xx10^(-34)Js`, Velocity of light: `c=3xx10^(8)m//s`]

Explanation

As `m`, `e`, `epsilon_(0)` and `h` are constants, we get
`E_(n)prop(1)/(n^(2))`
i.e., the energy of the electron in a stationery energy state is discrete and is inversely propotional to the square of the principal quantum number.
Nuclear binding energy : The minimum energy required to separate a nucleus into its free constituents i.e., protons and neutrons, is known as the nuclear binding energy.
Numerical :
Given : thresold wavelength `=230nm`
incident wavelength `=180nm`
`E_(max)=hc((1)/(lambda)-(1)/(lambda_(0)))`
`=6.63xx10^(-34)xx3xx10^(8)`
`((1)/(180xx10^(-9))-(1)/(230xx10^(-9)))`
`=6.63xx10^(-34)xx3xx10^(8)(5.56xx10^(6)-4.35xx10^(6))`
`=2.41xx10^(-19)J`
`=(2.41xx10^(-19))/(1.6xx10^(-19))`
`=1.5062eV`