What is the maximum wavelength line in the Ly | 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}} \

Vedclass · Accepted Answer

$\frac{1}{3R}$

Vedclass · Answer

(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} ight)$.For the Lyman series,$n_1 = 1$. For maximum wavelength,the transition must be from the nearest energy level,so $n_2 = 2$.For $He^{+}$ ion,the atomic number $Z = 2$.Substituting these values: $\frac{1}{\lambda_{\max}} = R (2)^{2} \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} ight)$.$\frac{1}{\lambda_{\max}} = R 	imes 4 	imes \left( 1 - \frac{1}{4} ight) = R 	imes 4 	imes \frac{3}{4} = 3R$.Therefore,$\lambda_{\max} = \frac{1}{3R}$.

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

Similar Questions

The wave number of the spectral line in the emission spectrum of hydrogen will be equal to $\frac{8}{9}$ times the Rydberg's constant if the electron jumps from

If the wavelength of light is $5.89 \times 10^{-5} \ cm$,then the energy of one photon in the given light will be ......

Calculate the energy associated with the second orbit of $Li^{+}$ and find its radius.

In the hydrogen spectrum,if an electron transitions from the $6^{th}$ orbit to the $3^{rd}$ orbit,what is the total number of spectral lines observed in the infrared region?

Which of the following results is not true about the photoelectric effect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module