Question

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

Answer 1

Answer:

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

The expression for Bohr velocity is:

$v=\frac{Ze^2}{2 \epsilon_0\times n\times h}$

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:

$v=\frac {2.185\times 10^6}{n}\ m/s$

Given, n = 2

So,

$v=\frac {2.185\times 10^6}{2}\ m/s$

$v=1.0925\times 10^6\ m/s$

Kinetic energy is:

$K.E.=\frac {1}{2}\times mv^2$

So,

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

K.E. = 5.4362 × 10⁻¹⁹ J