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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.

User Roshil K
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1 Answer

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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).

User Antiblank
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