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 $=136 eV )$

Where $R =$ Rydberg constant

$c =$ Speed of light in vacuum

$h =$ Planck's constant

• A. $40.8 \,eV$
• B. $3.4 veV$
• C. $10.5 \,eV$
• D. $13.6 \, eV$

Answer 1

Answer: (A)

$40.8 \,eV$

Answer 2

ΔE=13.6Z2[221​−421​]eV
=13.6×(4)2(41​−161​)eV
=13.6[4−1]eV
=13.6×3=40.8eV