A hydrogen atom in its ground state absorbs 10.2eV of energy. By what amount is the orbital angular momentum increased?

Question

A hydrogen atom in its ground state absorbs 10.2eV of energy. By what amount is the orbital angular momentum increased? (A) 1.05× 10- 34Jsec (B) 3.16× 10- 34Jsec (C) 2.11× 10- 34Jsec (D) 4.22× 10- 34Jsec

Tardigrade Admin · Accepted Answer

Energy of electron at ground state (n = 1) is -13.6eV after absorbing 10.2eV energy it total energy becomes -3.4eV . This is energy of first excited state (n = 2) So, Increase in momentum =(2 h/2 π )-(h/2 π )=(h/2 π ) =(6 . 6 × 10- 34/6 . 28)=1.05× 10- 34J-s

Q. A hydrogen atom in its ground state absorbs $10.2 e V$ of energy. By what amount is the orbital angular momentum increased?

$1.05 \times 1 0^{- 34} J sec$

$3.16 \times 1 0^{- 34} J sec$

$2.11 \times 1 0^{- 34} J sec$

$4.22 \times 1 0^{- 34} J sec$

Q. A hydrogen atom in its ground state absorbs 10.2eV of energy. By what amount is the orbital angular momentum increased?

1.05×10−34Jsec

3.16×10−34Jsec

2.11×10−34Jsec

4.22×10−34Jsec

Energy of electron at ground state (n=1) is −13.6eV after absorbing 10.2eV energy it total energy becomes −3.4eV . This is energy of first excited state (n=2) So, Increase in momentum =2π2h​−2πh​=2πh​ =6.286.6×10−34​=1.05×10−34J−s

Q. A hydrogen atom in its ground state absorbs $10.2 e V$ of energy. By what amount is the orbital angular momentum increased?

$1.05 \times 1 0^{- 34} J sec$

$3.16 \times 1 0^{- 34} J sec$

$2.11 \times 1 0^{- 34} J sec$

$4.22 \times 1 0^{- 34} J sec$