If the wavelengths of the first Lyman line fo | 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 a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$36:9:4$

Vedclass · Answer

(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the first Lyman line,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda} = R Z^2 \left( 1 - \frac{1}{4} \right) = R Z^2 \left( \frac{3}{4} \right)$.This implies $\lambda \propto \frac{1}{Z^2}$.For Hydrogen $(Z=1)$,$He^+$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$:$\lambda_1 : \lambda_2 : \lambda_3 = \frac{1}{1^2} : \frac{1}{2^2} : \frac{1}{3^2} = 1 : \frac{1}{4} : \frac{1}{9}$.Multiplying by $36$ to simplify the ratio:$\lambda_1 : \lambda_2 : \lambda_3 = 36 : 9 : 4$.

If the wavelengths of the first Lyman line for Hydrogen,$He^+$,and $Li^{2+}$ ions are $\lambda_1$,$\lambda_2$,and $\lambda_3$ respectively,then the ratio of these wavelengths is:

Similar Questions

Using the Rydberg formula,calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is

The shortest wavelength in the Balmer series of the hydrogen atom spectrum is approximately equal to (use $R_{H} = 1.097 \times 10^7 \ \text{m}^{-1}$) (in $\text{Å}$)

The shortest wavelength of the Lyman series of a hydrogen atom is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is equal to:

The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition of an electron from the:

Vedclass Test Series

Exam Paper Generator

Online Exam Module