The ionisation energy of a certain element is | vedclass

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

Question

(B) The energy difference between the ground state $(GS)$ and the first excited state $(1^{st} ES)$ is given by the difference in their respective ion

Vedclass · Accepted Answer

$UV$

Vedclass · Answer

(B) The energy difference between the ground state $(GS)$ and the first excited state $(1^{st} ES)$ is given by the difference in their respective ionisation energies: $\Delta E = IE_{GS} - IE_{1^{st} ES} = 1540 \ kJ \ mol^{-1} - 300 \ kJ \ mol^{-1} = 1240 \ kJ \ mol^{-1}$.To convert this energy into electron volts $(eV)$,we use the conversion factor $1 \ eV \approx 96.5 \ kJ \ mol^{-1}$:$\Delta E = \frac{1240}{96.5} \approx 12.85 \ eV$.Using the relation $\lambda (nm) = \frac{1240}{\Delta E (eV)}$:$\lambda = \frac{1240}{12.85} \approx 96.5 \ nm$.Since the wavelength is approximately $96.5 \ nm$,which falls in the range of $10 \ nm$ to $400 \ nm$,the emitted light lies in the $UV$ region.

Similar Questions

Find the energy of a photon having a wavelength of $6000 \ \mathring{A}$. (Given: $h = 6.62 \times 10^{-27} \ erg \cdot s$)

Which transition in a $H$ atom emits the maximum energy?

An electronic transition in a hydrogen atom results in the formation of the $H_\alpha$ line of the hydrogen spectrum in the Lyman series. The energies associated with the electron in each of the orbits involved in the transition (in $kcal \ mol^{-1}$) are:

The total energy of the electron of $H$ atom in the second quantum state is $-E_2$. The total energy of the $He^{+}$ ion in the third quantum state is

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $0.5\, V$ when the radiation of $250\, nm$ is used. The work function of the metal is ................ $eV$

Vedclass Test Series

Exam Paper Generator

Online Exam Module