When electronic transition occurs from a high | vedclass

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

Question

(A) The energy of a photon is given by $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength,we get $\lambda = \frac{hc}{\Delta E}$.Given $\Delta

Vedclass · Accepted Answer

$\frac{12395}{\Delta E} 	imes 10^{-10} \ m$

Vedclass · Answer

(A) The energy of a photon is given by $\Delta E = \frac{hc}{\lambda}$.Rearranging for wavelength,we get $\lambda = \frac{hc}{\Delta E}$.Given $\Delta E$ is in electron volts $(eV)$,we use $h = 6.626 	imes 10^{-34} \ J \ s$,$c = 3 	imes 10^{8} \ m/s$,and $1 \ eV = 1.602 	imes 10^{-19} \ J$.$\lambda = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^{8}}{\Delta E 	imes 1.602 	imes 10^{-19}} \ m$.$\lambda \approx \frac{12395}{\Delta E} 	imes 10^{-10} \ m$.

When electronic transition occurs from a higher energy state to a lower energy state with an energy difference equal to $\Delta E$ electron volts,the wavelength of the emitted line is approximately equal to

Similar Questions

Energy and radius of the first Bohr orbit of $He^{+}$ and $Li^{2+}$ are: $[$Given $R_{H} = 2.18 \times 10^{-18} \ J, a_{0} = 52.9 \ pm$ $]$

The ratio between the radii of $He^{+}$ ion and $H$ atom is:

The expression for the radius of the $n^{th}$ Bohr orbit of a hydrogen-like atom is:

Bohr's model cannot explain the emission spectrum of

The energy of an electron in the first Bohr orbit of hydrogen atom is $-2.18 \times 10^{-18} \ J$. Its energy in the third Bohr orbit is $........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module