Question

Ionisation energy of H atom in the ground state
is X KJ. Energy
required by an electron to jump from 2nd orbit to 3rd orbit will be.....? (I know the answer is 5x/36.) ONLY give answer with EXPLANATION.​

Answer 1

$0-\frac{-13.6}{1^2}$Please mark me brainliest

Here's the explanation:

$\frac{-E_0}{z^2}$ = E

Now, $E_0 = -13.6 eV$ = $-13.6x1.6x10^-^1^9 \frac{\frac{}{} }{} \frac{}{} \frac{}{}Joules$

Now, Difference in energy = $E_i_n_f_i_n_i_t_y-E_1$

Ionization energy from ground state = x KJ =  $\frac{-E_0}{infinity^2} - (\frac{-E_0}{1^2} )$= E₀ (Z=1, since atomic number of hydrogen is 1, and I've put the value zero, because anything upon infinity is zero)

Therefore, we will be determining the energy required to excite the electron from 2nd to 3rd state by 'x'

Now, Energy required to excite the electron from 2nd to 3rd state (or) orbit is:

ΔE = $E_3 - E_2$

=> $\frac{-x}{3^2} - (\frac{-x}{2^2} ) = \frac{x}{2^2} - \frac{x}{3^2}$ = $\frac{x}{4} -\frac{x}{9} = \frac{9x-4x}{36}$ = $\frac{5x}{36}$