What is the maximum wavelength in the Lyman s | vedclass

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

Question

(B) For the Lyman series,the transition occurs from higher energy levels to $n_1 = 1$. For maximum wavelength,the energy difference must be minimum,wh

Vedclass · Accepted Answer

$\frac{1}{3R}$

Vedclass · Answer

(B) For the Lyman series,the transition occurs from higher energy levels to $n_1 = 1$. For maximum wavelength,the energy difference must be minimum,which corresponds to the transition from $n_2 = 2$ to $n_1 = 1$. The Rydberg formula is $\frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$. For $He^+$ ion,$Z = 2$,so $Z^2 = 4$. Substituting the values: $\frac{1}{\lambda} = R 	imes 4 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 4R \left( 1 - \frac{1}{4} ight) = 4R \left( \frac{3}{4} ight) = 3R$. Therefore,$\lambda = \frac{1}{3R}$.

What is the maximum wavelength in the Lyman series of $He^{+}$ ion?

Similar Questions

For the hydrogen emission spectrum,the energy change $E$ (in Joules) is given by $\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ J$,where $n_1 = 1, 2, 3, \dots$ and $n_2 = 2, 3, 4, \dots$. Which of the following corresponds to the Paschen series?

Find the wave number of the shortest wavelength in the Balmer series.

What will be the frequency of the radiation emitted when an electron in a hydrogen atom jumps from $n = 4$ to $n = 1$? $(h = 6.625 \times 10^{-34} \,J \cdot s)$

The Ritz combination principle states that:

Which of the following transitions emits a photon with the maximum frequency?

Vedclass Test Series

Exam Paper Generator

Online Exam Module