Question

What is the value for the radius r for a n= 6 Bohr orbit electron in A (14 = 0.1 nm) Required precision = 2% Sanity check: answers should be between 0 and 20.

Answer 1

Step-by-step explanation:

It is known that
`mv_(r) = (nh)/(2p)`

where, m = mass of the electron

r = radius of the orbit

`v_(r)` = orbital speed of the electron

Equation when the electron is experiencing uniform circular motion is as follows.

`(Kze^(2))/(r^(2)) = (mv^(2))/(r)` ........ (1)

Rearranging above equation, we get the following.

`mv^(2) = (Kze^(2))/(r)`

Also, v =
`(nh)/(2pmr)` .......... (2)

Putting equation (2) in equation (1) we get the following.

`(mn^(2)h^(2))/(4p^(2)m^(2)r^(2)) = (Kze^(2))/(r)`

Hence, formula for radius of the nth orbital is as follows.

`r_(n) = [(h^(2))/(4p^(2)mKze^(2))]n^(2)`

`r_(n) = [5.29 * 10^(-11)m] * (6)^(2)`

=
`19.044 * 10^(-10) m`

=
`19.044 A^(o)`

Thus, we can conclude that the value for the radius r for a n= 6 Bohr orbit is
`19.044 A^(o)`.