In an emission spectrum,the transitions n_2 t | vedclass

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

Question

(B) The energy of a photon emitted during an electronic transition in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_{lower}^2} - \

Vedclass · Accepted Answer

$E_4 < E_3 < E_2 < E_1$

Vedclass · Answer

(B) The energy of a photon emitted during an electronic transition in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_{lower}^2} - \frac{1}{n_{higher}^2}) ext{ eV}$.As the principal quantum number $n$ increases,the energy gap between consecutive shells decreases.Therefore,the energy difference for the transitions are:$E_1 (n=2 o n=1) > E_2 (n=3 o n=2) > E_3 (n=4 o n=3) > E_4 (n=5 o n=4)$.Thus,the increasing order of energies is $E_4 < E_3 < E_2 < E_1$.

Metal	$Li$	$Na$	$K$	$Mg$	$Cu$	$Ag$
$W_0 / eV$	$2.42$	$2.3$	$2.25$	$3.7$	$4.8$	$4.3$

In an emission spectrum,the transitions $n_2 \to n_1$,$n_3 \to n_2$,$n_4 \to n_3$,and $n_5 \to n_4$ emit energies $E_1, E_2, E_3$,and $E_4$ respectively. What is the increasing order of these energies?

Similar Questions

The work function of a metal is $4.2 \, eV$. If radiation of $2000 \, \mathring{A}$ falls on the metal,then the kinetic energy of the fastest photoelectron is:

If the potential energy of the electron is $-10 \ eV$,what is the total energy in $eV$?

Calculate the energy of half a mole of photons of a radiation with a frequency of $3 \times 10^{12} \ Hz$.

The Ritz combination principle states that:

Vedclass Test Series

Exam Paper Generator

Online Exam Module