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

User Plinehan
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1 Answer

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Answer:

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

User KonstantinK
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