An atom absorbs a photon of wavelength 500,nm | vedclass

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

Question

(B) The net energy absorbed by the atom is given by the difference between the energy of the absorbed photon and the emitted photon.$E_{	ext{net}} =

Vedclass · Accepted Answer

$4125$

Vedclass · Answer

(B) The net energy absorbed by the atom is given by the difference between the energy of the absorbed photon and the emitted photon.$E_{	ext{net}} = E_{	ext{absorbed}} - E_{	ext{emitted}} = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} ight)$Substituting the given values:$E_{	ext{net}} = (6.6 	imes 10^{-34}) 	imes (3 	imes 10^8) \left( \frac{1}{500 	imes 10^{-9}} - \frac{1}{600 	imes 10^{-9}} ight)$$E_{	ext{net}} = 19.8 	imes 10^{-26} \left( \frac{600 - 500}{300000 	imes 10^{-18}} ight) = 19.8 	imes 10^{-26} \left( \frac{100}{3 	imes 10^{-13}} ight)$$E_{	ext{net}} = 19.8 	imes 10^{-26} 	imes 33.33 	imes 10^{13} = 6.6 	imes 10^{-20}\,J$To convert this energy into $eV$,divide by $1.6 	imes 10^{-19}\,J/eV$:$E_{	ext{net}} = \frac{6.6 	imes 10^{-20}}{1.6 	imes 10^{-19}} = 4.125 	imes 10^{-1}\,eV$$E_{	ext{net}} = 4125 	imes 10^{-4}\,eV$Comparing this with $n 	imes 10^{-4}\,eV$,we get $n = 4125$.

Similar Questions

When light of wavelength $\lambda$ is incident on a photosensitive surface,photons of power $P$ are emitted. The number of photons $n$ emitted in time $t$ is: [$h$ = Planck's constant,$c$ = velocity of light in vacuum]

When a photon enters glass from air,which one of the following quantities does $\underline{\text{NOT}}$ change?

The number of photons emitted by a $10\,W$ bulb in $10\,s,$ if the wavelength of light is $1000\,\mathring{A},$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module