Question

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rotates in a circular orbit about this nucleus. In the n = 5, orbit the electron is 1.32 10-9 m from the nucleus and it rotates with an angular speed of 3.30 1014 rad/s. Determine the electron's centripetal acceleration in m/s2.

Answer 1

Step-by-step explanation:

It is given that the atom is hydrogen. And, its electron rotates in n = 5 orbit with angular speed
`3.3 * 10^(14) rad/s`.

Radius of circular path =
`1.32 * 10^(-9)` m

Hence, first calculate the tangential speed as follows.

Tangential speed = radius × angular speed

=
`4.356 * 10^(5) m/s`

As formula to calculate time period is as follows.

T =
`1.5211 * 10^(-16) * (n^(3))/(z^(2))` sec

Also, frequency
`(\\u)` =
`(1)/(T)`

So, for n = 5 and z = 1 the value of frequency is as follows.

frequency
`(\\u)` =
`(1)/(1.5211 * 10^(-16)) * (z^(2))/(n^(3))`

=
`(1)/(1.5211 * 10^(-16)) * (1)/((5)^(3))`

=
`5.259 * 10^(13)` Hz

As formula to calculate centripetal acceleration is as follows.

`a_(c) = (v^(2))/(r)`

where, v = linear speed

r = radius

v =
`2.165 * 10^(6) * (z)/(n)`

=
`2.165 * 10^(6) * (1)/(5)`

=
`4.3 * 10^(5) m/s`

Hence, the centripetal acceleration will be as follows.

`a_(c) = (v^(2))/(r)`

=
`((4.3 * 10^(5))^(2))/(1.32 * 10^(-9))`

=
`1.4 * 10^(20) m/s^(2)`

Thus, we can conclude that the electron's centripetal acceleration is
`1.4 * 10^(20) m/s^(2)`.