An electron of a hydrogen-like atom,having Z= | vedclass

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

Question

(D) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.The energy rel

Vedclass · Accepted Answer

$40.8$

Vedclass · Answer

(D) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.The energy released when an electron jumps from an initial state $n_i$ to a final state $n_f$ is $\Delta E = E_{n_i} - E_{n_f} = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Given $Z = 4$,$n_i = 4$,and $n_f = 2$:$\Delta E = 13.6 	imes (4)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 16 \left( \frac{1}{4} - \frac{1}{16} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 16 \left( \frac{4-1}{16} ight) 	ext{ eV}$.$\Delta E = 13.6 	imes 3 	ext{ eV} = 40.8 	ext{ eV}$.

Similar Questions

If the ionization potential of an atom is $122.4 \, V$,then the excitation potentials for the first and second excited states are respectively:

The ionization potential of a hydrogen atom is $13.6 \text{ eV}$. How much energy needs to be supplied to ionize a hydrogen atom in the first excited state (in $\text{ eV}$)?

The energy levels of an atom are shown in the figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \, nm$? Given $(h = 6.62 \times 10^{-34} \, Js)$.

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively:

The ratio of the energies of the hydrogen atom in its first to second excited state is

Vedclass Test Series

Exam Paper Generator

Online Exam Module