An atom emits energy equal to 4 times 10^{-12 | vedclass

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

Question

(B) The wavelength $\lambda$ corresponding to energy $\Delta E = 4 	imes 10^{-12} \, erg$ is calculated using the formula $\lambda = \frac{hc}{\Delta

Vedclass · Accepted Answer

Visible region

Vedclass · Answer

(B) The wavelength $\lambda$ corresponding to energy $\Delta E = 4 	imes 10^{-12} \, erg$ is calculated using the formula $\lambda = \frac{hc}{\Delta E}$.Substituting the values: $\lambda = \frac{6.626 	imes 10^{-27} \, erg \cdot s 	imes 3 	imes 10^{10} \, cm/s}{4 	imes 10^{-12} \, erg}$.$\lambda = 4.9695 	imes 10^{-5} \, cm = 4969.5 \, \mathring{A}$.The visible region of the electromagnetic spectrum extends from approximately $3800 \, \mathring{A}$ to $7600 \, \mathring{A}$.Since $4969.5 \, \mathring{A}$ falls within this range,the radiation belongs to the visible region.

An atom emits energy equal to $4 \times 10^{-12} \, erg$. To which part of the electromagnetic spectrum does it belong?

Similar Questions

Calculate the frequency of visible light whose wavelength is $456 \, nm$.

In which type of radiation do the mass number and atomic number remain unchanged?

Calculate the wave number and frequency of orange radiation having wavelength $6300 \ \mathring{A}$.

Lines of the $Lyman$ series of the spectrum of hydrogen are in

$A$ photon of $300 \, nm$ is absorbed by a gas and then re-emits two photons. One re-emitted photon has a wavelength of $496 \, nm$. The wavelength of the second re-emitted photon is: (in $, nm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module