Question

The ionisation potential of hydrogen is 12.6
volt. Calculate the energy of its first excited
state.​

Answer 1

Given that,

Ionisation potential of hydrogen = -13.6 volt

Suppose, when the hydrogen atom jumps from first excited state to ground state.

n =2 to n = 1

We know that,

The energy state formula is defined as,

$E_{n}=\dfrac{-13.6}{n^2}$

Where, n = excited state

E = energy

We need to calculate the energy of its first excited state

Using formula of energy of hydrogen

$E_{n}=\dfrac{-13.6}{n^2}-\dfrac{-13.6}{n^2}$

Put the value into the formula

$E_{2}=\dfrac{-13.6}{2^2}-\dfrac{-13.6}{1^2}$

$E_{2}=\dfrac{-13.6}{4}-\dfrac{-13.6}{1}$

$E_{2}=10.2\ eV$

Hence, The energy of its first excited state is 10.2 eV.