A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (. Given, Plank's constant .=6.6 × 10-34 Js )

Question

A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (. Given, Plank's constant .=6.6 × 10-34 Js ) (A) 2.10 × 10-34 Js (B) 1.05 × 10-34 Js (C) 3.15 × 10-34 Js (D) 4.2 × 10-34 Js

Tardigrade Admin · Accepted Answer

13.6((1/12)-(1/n2))=10.2 n=2 Li=(h/2 π) × 1 LF=(2 h/2 π) Δ L=LF-Li=(h/2 π)=(6.6 × 10-34/22) =1.05 × 10-34 J-s

Q. A hydrogen atom in is ground state absorbs $10.2 e V$ of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : $($ Given, Plank's constant $= 6.6 \times 1 0^{- 34} J s)$

$2.10 \times 1 0^{- 34} J s$

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

$3.15 \times 1 0^{- 34} J s$

$4.2 \times 1 0^{- 34} J s$

$13.6 (\frac{1}{1 ^{2}} - \frac{1}{n ^{2}}) = 10.2$
$n = 2$
$L_{i} = \frac{h}{2 π} \times 1$
$L_{F} = \frac{2 h}{2 π}$
$Δ L = L_{F} - L_{i} = \frac{h}{2 π} = \frac{6.6 \times 1 0 ^{- 34}}{22}$
$= 1.05 \times 1 0^{- 34} J - s$

Q. A hydrogen atom in is ground state absorbs 10.2eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : ( Given, Plank's constant =6.6×10−34Js)

2.10×10−34Js

1.05×10−34Js

3.15×10−34Js

4.2×10−34Js