If the series limit of the Lyman series for a | vedclass

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

Question

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the series limit of the Lyman series of a Hydrogen atom $(Z=1)$,we have $n_1 = 1$ and $n_2 = \infty$:$\frac{1}{\lambda_L} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty} \right] = R$.Thus,$\lambda_L = \frac{1}{R}$.For the series limit of the Balmer series of a hydrogen-like atom (atomic number $Z$),we have $n_1 = 2$ and $n_2 = \infty$:$\frac{1}{\lambda_B} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty} \right] = \frac{R Z^2}{4}$.Thus,$\lambda_B = \frac{4}{R Z^2}$.Given that the series limits are equal,$\lambda_L = \lambda_B$:$\frac{1}{R} = \frac{4}{R Z^2} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

If the series limit of the Lyman series for a Hydrogen atom is equal to the series limit of the Balmer series for a hydrogen-like atom,then the atomic number of this hydrogen-like atom will be:

Similar Questions

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

The ratio of minimum wavelength of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :-

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is

The wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module