The smallest wavelength of the Lyman series i | vedclass

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

Question

(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Lyman series,the smallest wavelen

Vedclass · Accepted Answer

$1217$

Vedclass · Answer

(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,the smallest wavelength corresponds to $n_1 = 1$ and $n_2 = \infty$.$\frac{1}{\lambda_{L,min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R = \frac{1}{91} \ nm^{-1}$.For the Balmer series,the largest wavelength corresponds to $n_1 = 2$ and $n_2 = 3$.$\frac{1}{\lambda_{B,max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = \frac{1}{91} \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{1}{91} \left( \frac{5}{36} ight)$.$\lambda_{B,max} = \frac{91 	imes 36}{5} = 655.2 \ nm$.For the Paschen series,the largest wavelength corresponds to $n_1 = 3$ and $n_2 = 4$.$\frac{1}{\lambda_{P,max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = \frac{1}{91} \left( \frac{1}{9} - \frac{1}{16} ight) = \frac{1}{91} \left( \frac{7}{144} ight)$.$\lambda_{P,max} = \frac{91 	imes 144}{7} = 1872 \ nm$.The difference is $\Delta\lambda = \lambda_{P,max} - \lambda_{B,max} = 1872 - 655.2 = 1216.8 \ nm \approx 1217 \ nm$.

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

Similar Questions

In a hydrogen atom,the wavelength of the light emitted during a transition from the $n = 3$ orbit to the $n = 2$ orbit is ${\lambda _0}$. What will be the wavelength of the light emitted during a transition from the $n = 4$ orbit to the $n = 2$ orbit?

If the first line of Lyman series has a wavelength $1215.4 \text{ Å}$, the first line of Balmer series is approximately (in $\text{ Å}$)

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

The number of possible spectral lines in a hydrogen atom is:

The ratio of the maximum wavelengths of the Lyman and Balmer series in the hydrogen spectrum is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module