An electron of a hydrogen like atom, having Z=4, jumps from 4 text th energy state to 2 text nd energy state. The energy released in this process, will be: ( Given Rch =13.6 eV ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant

Question

An electron of a hydrogen like atom, having Z=4, jumps from 4 text th energy state to 2 text nd energy state. The energy released in this process, will be: ( Given Rch =13.6 eV ) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant (A) 40.8 eV (B) 3.4 eV (C) 10.5 eV (D) 13.6 eV

Tardigrade Admin · Accepted Answer

Δ E =13.6 Z 2[(1/22)-(1/42)] eV =13.6 ×(4)2((1/4)-(1/16)) eV =13.6[4-1] eV =13.6 × 3=40.8 eV

Q. An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be :
$($ Given Rch $= 13.6 e V)$
Where $R =$ Rydberg constant
$c =$ Speed of light in vacuum
$h =$ Planck's constant

$40.8 e V$

$3.4 e V$

$10.5 e V$

$13.6 e V$

$Δ E = 13.6 Z^{2} [\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}] e V$
$= 13.6 \times (4)^{2} (\frac{1}{4} - \frac{1}{16}) e V$
$= 13.6 [4 - 1] e V$
$= 13.6 \times 3 = 40.8 e V$

Q. An electron of a hydrogen like atom, having Z=4, jumps from 4th energy state to 2nd energy state. The energy released in this process, will be : ( Given Rch =13.6eV) Where R= Rydberg constant c= Speed of light in vacuum h= Planck's constant

40.8eV

3.4eV

10.5eV

13.6eV