In hydrogen spectrum, a hydrogen atom emits a photon of wavelength 1027 mathringA . Its angular momentum changes by

Question

In hydrogen spectrum, a hydrogen atom emits a photon of wavelength 1027 mathringA . Its angular momentum changes by (A) (h/π ) (B) (h/2 π ) (C) (3 h/2 π ) (D) (2 h/π )

Tardigrade Admin · Accepted Answer

The energy of the photon emitted E=(12430/1027)=12.1 eV This corresponds to transition from n=3 with E3=-1.5 eV to nf=1 with E1=-13.6 eV Hence Δ L=3((h/2 π ))-1((h/2 π ))=(h/π )

Q. In hydrogen spectrum, a hydrogen atom emits a photon of wavelength $1027 \overset{˚}{A}$ . Its angular momentum changes by

$\frac{h}{π}$

$\frac{h}{2 π}$

$\frac{3 h}{2 π}$

$\frac{2 h}{π}$

The energy of the photon emitted $E = \frac{12430}{1027} = 12.1 e V$
This corresponds to transition from
$n = 3$ with $E_{3} = - 1.5 e V$ to $n_{f} = 1$
with $E_{1} = - 13.6 e V$
Hence $Δ L = 3 (\frac{h}{2 π}) - 1 (\frac{h}{2 π}) = \frac{h}{π}$

Q. In hydrogen spectrum, a hydrogen atom emits a photon of wavelength 1027A˚ . Its angular momentum changes by

πh​

2πh​

2π3h​

π2h​