Atomic Physics Question 11
Question 11 - 01 February - Shift 2
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
(1) $13.6 eV$
(2) $10.5 eV$
(3) $3.4 eV$
(4) $40.8 eV$
Show Answer
Answer: (4)
Solution:
Formula: Wavelength Corresponding To Spectral Lines
$ \begin{aligned} & \Delta E=13.6 Z^{2}[\frac{1}{2^{2}}-\frac{1}{4^{2}}] eV \\ & =13.6 \times(4)^{2}(\frac{1}{4}-\frac{1}{16}) eV \\ & =13.6[4-1] eV \\ & =13.6 \times 3=40.8 eV \end{aligned} $
