Question

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?

Answer 1

K.E. = 5.4362 × 10⁻¹⁹ J

Step-by-step explanation:

The expression for Bohr velocity is:

`v=(Ze^2)/(2 \epsilon_0* n* h)`

Applying values for hydrogen atom,

Z = 1

Mass of the electron (
`m_e`) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

`\epsilon_0` = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

`v=\frac {2.185* 10^6}{n}\ m/s`

Given, n = 2

So,

`v=\frac {2.185* 10^6}{2}\ m/s`

`v=1.0925* 10^6\ m/s`

Kinetic energy is:

`K.E.=\frac {1}{2}* mv^2`

So,

`K.E.=\frac {1}{2}* 9.1093* 10^(-31)* ({1.0925* 10^6})^2`

K.E. = 5.4362 × 10⁻¹⁹ J