The electronic transitions from n=2 to n=1 wi | vedclass

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

Question

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

Vedclass · Accepted Answer

$Li^{2+}$

Vedclass · Answer

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ For the transition from $n=2$ to $n=1$: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = Z^2 \cdot R_H \left( \frac{3}{4} \right)$ This implies $\lambda = \frac{4}{3 Z^2 R_H}$, so $\lambda \propto \frac{1}{Z^2}$. To obtain the shortest wavelength, the atomic number $Z$ must be the maximum. Comparing the species: $H$ $(Z=1)$, $He^{+}$ $(Z=2)$, and $Li^{2+}$ $(Z=3)$. Since $Li^{2+}$ has the highest $Z$, it will produce the shortest wavelength.

The electronic transitions from $n=2$ to $n=1$ will produce the shortest wavelength in which of the following species? (where $n$ is the principal quantum state)

Similar Questions

The wavelength of the first line of the Balmer series in the hydrogen spectrum is ................ $\mathring{A}$.

The energy required to excite an electron from the first orbit to the third orbit in a hydrogen atom is .......... $eV$.

State the uses of atomic spectra.

If the energies of two light radiations $E_1$ and $E_2$ are $25 \ eV$ and $100 \ eV$ respectively,then their respective wavelengths $\lambda_1$ and $\lambda_2$ would be in the ratio $\lambda_1: \lambda_2=$

If the wavelength of an electromagnetic radiation is $2000 \ \text{Å}$, what is its energy in ergs?

Vedclass Test Series

Exam Paper Generator

Online Exam Module