The energy required (in eV) to excite an elec | vedclass

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

Question

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

Vedclass · Accepted Answer

$12.1$

Vedclass · Answer

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = \frac{-13.6 \ Z^2}{n^2} \ eV$. For the ground state $(n=1)$ of $H$-atom $(Z=1)$: $E_1 = \frac{-13.6 	imes 1^2}{1^2} = -13.6 \ eV$. The third excited state corresponds to $n=4$ (since $n=1$ is ground,$n=2$ is first excited,$n=3$ is second excited,and $n=4$ is third excited). $E_4 = \frac{-13.6 	imes 1^2}{4^2} = \frac{-13.6}{16} = -0.85 \ eV$. The energy required for excitation is $\Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \ eV$. Note: If the question implies the third state $(n=3)$,then $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \ eV \approx 12.1 \ eV$. Given the options,the question refers to the third state $(n=3)$.

The energy required (in $eV$) to excite an electron of $H$-atom from the ground state to the third excited state is

Similar Questions

What is the energy associated with the first orbit of $Li^{2+}$ $(R_{H} = 2.18 \times 10^{-18} \ J)$?

Evaluate the following ratios for the energy of the electron in a particular orbit:
[Kinetic $:$ Potential] and [Total $:$ Kinetic]

If the shortest wavelength of the $He^{+}$ ion in the Balmer series is $X \ m$,then the longest wavelength in the Paschen series of the $Li^{2+}$ ion is:

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

Which of the following electronic transitions in a hydrogen atom results in the emission of a spectral line with the shortest wavelength?

Vedclass Test Series

Exam Paper Generator

Online Exam Module