The wavelength of the radiation emitted,when | 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 emitted radiation is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_

Vedclass · Accepted Answer

$91$

Vedclass · Answer

(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$For a hydrogen atom,$Z = 1$. The electron falls from infinity $(n_2 = \infty)$ to the stationary state $n_1 = 1$.Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \ m^{-1} 	imes (1)^2 	imes (\frac{1}{1^2} - \frac{1}{\infty^2})$$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes (1 - 0) = 1.097 	imes 10^7 \ m^{-1}$$\lambda = \frac{1}{1.097 	imes 10^7} \ m \approx 9.115 	imes 10^{-8} \ m$To convert meters to nanometers $(nm)$,multiply by $10^9$: $\lambda = 9.115 	imes 10^{-8} 	imes 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

The wavelength of the radiation emitted,when in a hydrogen atom an electron falls from infinity to stationary state $n = 1$,would be ........ $nm$ (Rydberg constant $R_H = 1.097 \times 10^7 \ m^{-1}$)

Similar Questions

The first quantum mechanical model of the atom was proposed by ........

If the shortest wavelength of the hydrogen atom in the Lyman series is $x$, then the longest wavelength in the Balmer series of $He^+$ is:

The ratio of radii of the second orbit of a hydrogen atom to the fourth orbit of a $He^{+}$ ion is:

The radii of the first Bohr orbit of $H$ $(r_{H})$,$He^{+}$ $(r_{He^{+}})$ and $Li^{2+}$ $(r_{Li^{2+}})$ are in the order

The ratio of kinetic energies of photoelectrons emitted from a metal by light of frequencies $v_1$ and $v_2$ is $1 : 2$. The threshold frequency of the metal will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module