Transitions between three energy levels in a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) From the energy level diagram,the energy of the emitted photon is given by $\Delta E = \frac{hc}{\lambda}$.For the transitions shown:$1$) Transiti

Vedclass · Accepted Answer

$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

Vedclass · Answer

(C) From the energy level diagram,the energy of the emitted photon is given by $\Delta E = \frac{hc}{\lambda}$.For the transitions shown:$1$) Transition from $E_3$ to $E_1$ corresponds to wavelength $\lambda_1$: $E_3 - E_1 = \frac{hc}{\lambda_1} \quad (1)$$2$) Transition from $E_2$ to $E_1$ corresponds to wavelength $\lambda_2$: $E_2 - E_1 = \frac{hc}{\lambda_2} \quad (2)$$3$) Transition from $E_3$ to $E_2$ corresponds to wavelength $\lambda_3$: $E_3 - E_2 = \frac{hc}{\lambda_3} \quad (3)$We observe that $(E_3 - E_1) = (E_3 - E_2) + (E_2 - E_1)$.Substituting the expressions from equations $(1), (2),$ and $(3)$:$\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$Dividing both sides by $hc$,we get:$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

Similar Questions

As the quantum number $n$ increases,the difference in energy between consecutive energy levels

The minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is nearly: (in $eV$)

In a hydrogen atom,if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E,$ the ionization energy of the hydrogen atom is ....... $E$.

The number of spectral lines emitted by atomic hydrogen that is in the $4^{\text{th}}$ energy level is

Vedclass Test Series

Exam Paper Generator

Online Exam Module