When a hydrogen atom emits a photon of energy 12.09 eV, its orbital angular momentum changes by (where h is Planck's constant)

Question

When a hydrogen atom emits a photon of energy 12.09 eV, its orbital angular momentum changes by (where h is Planck's constant) (A) (3 h/π) (B) (2 h/π) (C) (h/π) (D) (4 h/π)

Tardigrade Admin · Accepted Answer

This is the case of electron jumping from 3 text rd orbit to 115 t orbit Δ L =(n1 h/2 π)-(n h/2 π) =(3 h/2 π)-(h/2 π)=(h/π)

Q. When a hydrogen atom emits a photon of energy $12.09 e V$ , its orbital angular momentum changes by (where $h$ is Planck's constant)

$\frac{3 h}{π}$

$\frac{2 h}{π}$

$\frac{h}{π}$

$\frac{4 h}{π}$

This is the case of electron jumping from $3^{rd}$ orbit to $1^{15 t}$ orbit
$Δ L = \frac{n _{1} h}{2 π} - \frac{nh}{2 π}$
$= \frac{3 h}{2 π} - \frac{h}{2 π} = \frac{h}{π}$

Q. When a hydrogen atom emits a photon of energy 12.09eV, its orbital angular momentum changes by (where h is Planck's constant)

π3h​

π2h​

πh​

π4h​