Energy E of a hydrogen atom with principal quantum number n is given by E = (-13.6/n2) eV The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :-

Question

Energy E of a hydrogen atom with principal quantum number n is given by E = (-13.6/n2) eV The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :- (A) 0. 85 eV (B) 3.4 eV (C) 1.9 eV (D) 1.5 eV

Tardigrade Admin · Accepted Answer

Energy of photon = E3 - E2 = - (-13.6/9) - ( ( - 13.6/4)) = (5/36) × 13.6 = 1.9 eV

Q. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{- 13.6}{n ^{2}} e V$
The energy of a photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately :-

0. 85 eV

3.4 eV

1.9 eV

1.5 eV

Energy of photon = $E_{3} - E_{2}$
$= - \frac{- 13.6}{9} - (\frac{- 13.6}{4}) = \frac{5}{36} \times 13.6 = 1.9 e V$

Q. Energy E of a hydrogen atom with principal quantum number n is given by E=n2−13.6​eV The energy of a photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately :-