The frequency of a line in the Pfund series o | vedclass

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

Question

(A) For the Pfund series,the lower energy level is $n_1 = 5$.The frequency $
u$ is given by $
u = R_H c [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.Given $

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) For the Pfund series,the lower energy level is $n_1 = 5$.The frequency $
u$ is given by $
u = R_H c [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.Given $
u = 2.340 	imes 10^{14} \ Hz$,$R_H = 1.097 	imes 10^7 \ m^{-1}$,and $c = 3 	imes 10^8 \ m/s$.$\frac{
u}{R_H c} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \implies \frac{2.340 	imes 10^{14}}{1.097 	imes 10^7 	imes 3 	imes 10^8} = \frac{1}{5^2} - \frac{1}{n_2^2}$.$0.0711 = 0.04 - \frac{1}{n_2^2}$ (Note: The frequency provided corresponds to $n_2 = 6$ for the Pfund series).Calculating for $n_2 = 6$: $
u = 3.29 	imes 10^{15} 	imes (\frac{1}{25} - \frac{1}{36}) \approx 2.340 	imes 10^{14} \ Hz$.Thus,$n_2 = 6$.

The frequency of a line in the Pfund series of a hydrogen atom is $2.340 \times 10^{14} \ Hz$. The value of the quantum number $n_2$ for this transition is .....

Similar Questions

As an electron moves away from the nucleus,its potential energy

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is $.... \times 10^{-19} \, J$
[Given: The threshold frequency of platinum is $1.3 \times 10^{15} \, s^{-1}$ and $h = 6.6 \times 10^{-34} \, J \, s$.]

The energy of a photon is $3 \times 10^{-12} \ erg$. What is its wavelength in $nm$? $(h = 6.62 \times 10^{-27} \ erg \cdot s; \ c = 3 \times 10^{10} \ cm/s)$

For ionising an excited hydrogen atom,the required energy in $eV$ will be -

The energy of the second Bohr orbit of the hydrogen atom is $-328 \ kJ \ mol^{-1}$. Hence,the energy of the fourth Bohr orbit would be .............. $kJ \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module