Which statement relating to the spectrum of H | vedclass

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

Question

(C) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.As the wave

Vedclass · Accepted Answer

The spectral lines are closer together at longer wavelength

Vedclass · Answer

(C) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.As the wavelength $\lambda$ increases,the energy difference $\Delta E = \frac{hc}{\lambda}$ decreases.In any spectral series,as we move to higher energy levels (higher $n$),the energy gap between successive levels decreases,meaning the spectral lines become closer together at shorter wavelengths (higher energy),not longer wavelengths.Therefore,the statement that spectral lines are closer together at longer wavelengths is false.

Which statement relating to the spectrum of $H$ atom is false:

Similar Questions

Calculate the radius of the first orbit of $Li^{2+}$. (in $pm$)

When $\alpha$-particles are sent through a thin metal foil,most of them go straight through the foil because

The Bohr orbit radius for the hydrogen atom $(n = 1)$ is approximately $0.530 \ \mathring{A}$. The radius for the first excited state $(n = 2)$ orbit is ........... $\mathring{A}$

For the Paschen series,the values of $n_1$ and $n_2$ in the expression $\Delta E = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ are:

The wavelength of a spectral line for an electronic transition is inversely related to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module