Question

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a centripetal acceleration of 9.0 x 1022 m/s2.

a) Find its speed.
b) In an excited state with principle quantum number n = 10, the electron is 100 times farther away, and its centripetal acceleration is 10,000 times.

Answer 1

Step-by-step explanation:

Given

radius of electron
`(r)=5.3* 10^(-11) m`

centripetal acceleration
`(a_c)=9* 10^22 m/s^2`

we know

`a_c=(v^2)/(r)`

`v=√(r* a_c)`

`v=\sqrt{5.3* 10^(-11)* 9* 10^(22)}`

`v=√(47.7* 10^11)`

`v=21.84* 10^5 m/s`

(b)For n=10

`r=100* 5.3* 10^(-11) m\approx 5.3* 10^(-9) m`

`a_c=10^4* 9* 10^(22) m/s^2`

`a_c=9* 10^(26) m/s^2`

`v=√(r* a_c)`

`v=\sqrt{9* 10^(26)* 5.3* 10^(-9)}`

`v=21.84* 10^8 m/s`