Radius and Energy levels of hydrogen atom :
Consider an electron of mass 'm' and charge 'e' revolving around a nucleus of charge `Ze` (where, `Z` = atomic number and `e` is the charge of the proton) with a tangential velocity `v`. `r` is the radius of the orbit in which electron is revolving.
By Coulomb's Law, the electrostatic force of attraction between the moving electron and nucleus is Coulombic
force= `(KZe^2)/r^2`
`K = 1/(4 pi epsilon_0)` (where `epsilon_0` is permittivity of free space); `K = 9 xx 10^9` `N``m^2``C^(-2)`
In C.G.S. units, value of `K = 1` `dyn e` `cm^2` `(esu)^( -2)`
The centrifugal force acting on the electron is `(mv^2)/r`
Since the electrostatic force balance the centrifugal force, for the stable electron orbit.
`(mv^2)/r` = `(KZe^2)/r^2` or `v^2` = `(KZe^2)/r^2` ..........(2)
According to Bohr's postulate of angular momentum quantization, we have
`mvr = (nh)/(2 pi)`; `v = (nh)/(2 pi mr)`; `v^2 = (n^2h^2)/(4 pi^2 m^2 r^2)` ............(3)
Equating (2) and (3) `(KZe^2)/r^2` = `(n^2h^2)/(4 pi^2 m^2 r^2)`
SoIving for `r` we get `r` =`(n^2h^2)/(4 pi^2 m KZe^2)`
where `n` = `1`, `2`, `3`........
Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of `n`, i.e., farther the energy level from the nucleus, the greater is the radius.
The radius of the smallest orbit (`n=1`) for hydrogen atom (`Z=1`) is `r_0`.
`r_0` = `(n^2h^2)/(4 pi^2 m Ke^2)`
`= (1^2 xx(6.626 xx 10^(-34))^2)/(4xx(3.14)^2 xx9xx 10^(-31) xx(1.6 xx 10^(-19))^2 xx 9 xx 10^9)`
`= 5.29 xx 10^(-11)` m = `0.529` `A^o`
Radius of `n^(th)` orbit for an atom with atomic number `Z` is simply written as
`r_n = 0.529 xx (n^2)/Z` `A^o`
By Coulomb's Law, the electrostatic force of attraction between the moving electron and nucleus is Coulombic
force= `(KZe^2)/r^2`
`K = 1/(4 pi epsilon_0)` (where `epsilon_0` is permittivity of free space); `K = 9 xx 10^9` `N``m^2``C^(-2)`
In C.G.S. units, value of `K = 1` `dyn e` `cm^2` `(esu)^( -2)`
The centrifugal force acting on the electron is `(mv^2)/r`
Since the electrostatic force balance the centrifugal force, for the stable electron orbit.
`(mv^2)/r` = `(KZe^2)/r^2` or `v^2` = `(KZe^2)/r^2` ..........(2)
According to Bohr's postulate of angular momentum quantization, we have
`mvr = (nh)/(2 pi)`; `v = (nh)/(2 pi mr)`; `v^2 = (n^2h^2)/(4 pi^2 m^2 r^2)` ............(3)
Equating (2) and (3) `(KZe^2)/r^2` = `(n^2h^2)/(4 pi^2 m^2 r^2)`
SoIving for `r` we get `r` =`(n^2h^2)/(4 pi^2 m KZe^2)`
where `n` = `1`, `2`, `3`........
Hence only certain orbits whose radii are given by the above equation are available for the electron. The greater the value of `n`, i.e., farther the energy level from the nucleus, the greater is the radius.
The radius of the smallest orbit (`n=1`) for hydrogen atom (`Z=1`) is `r_0`.
`r_0` = `(n^2h^2)/(4 pi^2 m Ke^2)`
`= (1^2 xx(6.626 xx 10^(-34))^2)/(4xx(3.14)^2 xx9xx 10^(-31) xx(1.6 xx 10^(-19))^2 xx 9 xx 10^9)`
`= 5.29 xx 10^(-11)` m = `0.529` `A^o`
Radius of `n^(th)` orbit for an atom with atomic number `Z` is simply written as
`r_n = 0.529 xx (n^2)/Z` `A^o`