The ionization energy of the electron in the | vedclass

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

Question

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac

Vedclass · Accepted Answer

$n=4 	o n=3$

Vedclass · Answer

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac{n(n-1)}{2}$.Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$,so $n = 4$.The energy of the emitted photon is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is maximum when the energy difference $\Delta E$ is minimum.The possible transitions from $n=4$ are: $(4 	o 3), (4 	o 2), (4 	o 1), (3 	o 2), (3 	o 1), (2 	o 1)$.Comparing the energy gaps,the transition $n=4 	o n=3$ has the smallest energy difference,and therefore corresponds to the maximum wavelength of the emitted radiation.

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \text{ eV}$. The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. The maximum wavelength of the emitted radiation corresponds to the transition between:

Similar Questions

How many spectral lines are obtained when an electron in the ground state of hydrogen is excited to the principal quantum number $n = 3$?

When the electron in a hydrogen atom jumps from the fourth Bohr orbit to the second Bohr orbit,one gets the:

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively:

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module