Question

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

Answer 1

The kinetic energy is 86.6 zepto joules.

Step-by-step explanation:

Given that,

Number of orbit n =5

We know that,

Bohr's radius for hydrogen atom is

`r = 0.53*10^(-10)* n^2\ m`

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

`r=0.53*10^(-10)*5^2`

`r =1.33*10^(-9)\ m`

We need to calculate the kinetic energy

Using formula of kinetic energy

`E_(k)=(1)/(4\pi\epsilon_(0))*(e^2)/(2* r_(s))`

Put the value into the formula

`E_(k)=(9*10^(9)*(1.6*10^(-19))^2)/(2*1.33*10^(-9))`

`E_(k)=8.66*10^(-20)\ J`

We know that,

`1\ zepto\ joule=1*10^(-21)\ J`

The kinetic energy is

`E_(k)=(8.66*10^(-20))/(1*10^(-21))`

`E_(k)=86.6\ zepto\ Joules`

Hence, The kinetic energy is 86.6 zepto joules.