If R is the Rydberg constant in cm^{-1}, then | vedclass

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

Question

(B) The wavelength $\lambda$ of radiation emitted by a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} -

Vedclass · Accepted Answer

$\frac{7}{5R}$ to $\frac{19}{5R} \ cm$

Vedclass · Answer

(B) The wavelength $\lambda$ of radiation emitted by a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.For each spectral series, the range of wavelengths is determined by the transition from $n_i = n_f + 1$ to $n_i = \infty$.$1$. Lyman series $(n_f = 1)$: $\lambda$ ranges from $\frac{1}{R}$ to $\frac{4}{3R}$.$2$. Balmer series $(n_f = 2)$: $\lambda$ ranges from $\frac{4}{R}$ to $\frac{36}{5R}$.$3$. Paschen series $(n_f = 3)$: $\lambda$ ranges from $\frac{9}{R}$ to $\frac{144}{7R}$.Comparing the given options, the range $\frac{7}{5R}$ to $\frac{19}{5R}$ does not correspond to any of the spectral series of the hydrogen atom.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

If $R$ is the Rydberg constant in $cm^{-1}$, then the hydrogen atom does not emit any radiation of wavelength in the range of

Similar Questions

Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.

In the Lyman series,the series limit of wavelength is $\lambda_1$. The wavelength of the first line of the Lyman series is $\lambda_2$ and in the Balmer series,the series limit of wavelength is $\lambda_3$. Then the relation between $\lambda_1$,$\lambda_2$,and $\lambda_3$ is

For a hydrogen atom,$\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions $1$ and $2$ respectively,as shown in the figure. The ratio of $\lambda_1$ and $\lambda_2$ is $\frac{x}{32}$. The value of $x$ is:

The transition from the state $n = 3$ to $n = 1$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

Vedclass Test Series

Exam Paper Generator

Online Exam Module