Question

The kinetic energy for the H atom 13.6 eV then the required energy to excite it from the ground state to next higher state will be?

Answer 1

Answer:

According to Bohr, the energy possessed by an electron in each orbit in general is given by the equation:

Energy = -13.6 ev / n²

Here n refers to the number of orbit.

Here we are required to find the energy at the second orbit. So n = 2.

=> E = -13.6 ev / 2²

=> E = -13.6 ev / 4

=> E = - 3.4 ev

Hence this is the energy possessed by electron in second orbit. But energy required can be calculated as:

Energy in First orbit + Energy required = Energy in Second orbit

=> Energy Required = Energy in Second Orbit - Energy in first orbit

=> Energy Required = - 3.4 ev - ( - 13.6 ev )

=> Energy Required = - 3.4 + 13.6 = 10.2 ev

Hence the kinetic Energy required is 10.2 ev.

Answer 2

$\huge\ {Hey \:mate}$

$<b>✨ Here's ur answer<b>$

According to the question, the Ionisation potential of hydrogen atom is 13.6eV.

Therefore, the change in energy involved in removing the electron from n=2 is given by

the formula E(n)=−13.6n2 eV/atom

However, in this case the electron is completely removed from n=2, therefore

ΔE=E(∞)−E(2)

Therefore,

ΔE=0−(−13.622)

ΔE=+13.64

ΔE=3.4eV

Now.. The energy required to excite it is...

3.4-(13.6) v

$<u>Answer =10.2v</u >$


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