A hydrogen atom in the ground state is given | vedclass

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

Question

(D) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$.When an energy of $10.2 \ eV$ is provided,the new energy level $E_n$ is gi

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$.When an energy of $10.2 \ eV$ is provided,the new energy level $E_n$ is given by $E_n = E_1 + 10.2 \ eV = -13.6 \ eV + 10.2 \ eV = -3.4 \ eV$.Since $E_n = -13.6/n^2 \ eV$,we have $-3.4 = -13.6/n^2$,which gives $n^2 = 4$,so $n = 2$.The electron is excited to the first excited state $(n = 2)$.The number of spectral lines emitted when an electron transitions from state $n$ to the ground state is given by the formula $N = n(n-1)/2$.For $n = 2$,$N = 2(2-1)/2 = 1$.Therefore,only $1$ spectral line will be emitted.

$A$ hydrogen atom in the ground state is given an energy of $10.2 \ eV$. How many spectral lines will be emitted due to the transition of electrons?

Similar Questions

If $R$ is the Rydberg constant for hydrogen,the wave number of the first line in the Lyman series will be

Which one of the series of hydrogen spectrum is in the visible region?

The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra lies in the visible region?

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.
Reason : $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module