The electron in a hydrogen atom is initially | vedclass

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

Question

(B) The ground state corresponds to $n = 1$.The first excited state corresponds to $n = 2$.The second excited state corresponds to $n = 3$.When the el

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The ground state corresponds to $n = 1$.The first excited state corresponds to $n = 2$.The second excited state corresponds to $n = 3$.When the electron transitions from an initial state $n_i = 3$ to a final state $n_f = 1$, the number of possible spectral lines is given by the formula $N = \frac{n(n-1)}{2}$, where $n$ is the initial energy level.Alternatively, we can list the possible transitions: $1$. From $n = 3$ to $n = 2$$2$. From $n = 2$ to $n = 1$$3$. From $n = 3$ to $n = 1$Total number of spectral lines = $3$.

The electron in a hydrogen atom is initially in the second excited state. When it finally moves to the ground state, the maximum number of spectral lines emitted are:

Similar Questions

$v_{1}$ is the frequency of the series limit of the Lyman series,$v_{2}$ is the frequency of the first line of the Lyman series,and $v_{3}$ is the frequency of the series limit of the Balmer series. Then:

The colour of the second line of the Balmer series is

The ratio of minimum wavelengths of Lyman and Balmer series will be

$A$ $12.5\; eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

If Rydberg's constant is $R$,the longest wavelength of radiation in Paschen series will be $\frac{\alpha}{7 R}$,where $\alpha=$ . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module