Question

What is the value for the kinetic energy for a n = 5 Bohr orbit electron in zeptoJoules?

Answer 1

Answer:

The kinetic energy is 86.6 zepto joules.

Explanation:

Given that,

Number of orbit n =5

We know that,

Bohr's radius for hydrogen atom is

$r = 0.53\times10^{-10}\times n^2\ m$

Now, put the value of n in the formula of radius

$r=0.53\times10^{-10}\times5^2$

$r =1.33\times10^{-9}\ m$

We need to calculate the kinetic energy

Using formula of kinetic energy

$E_{k}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{e^2}{2\times r_{s}}$

Put the value into the formula

$E_{k}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times1.33\times10^{-9}}$

$E_{k}=8.66\times10^{-20}\ J$

We know that,

$1\ zepto\ joule=1\times10^{-21}\ J$

The kinetic energy is

$E_{k}=\dfrac{8.66\times10^{-20}}{1\times10^{-21}}$

$E_{k}=86.6\ zepto\ Joules$

Hence, The kinetic energy is 86.6 zepto joules.