Identify the pair of species having the same | vedclass

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

Question

(B) For a single-electron species, the energy is given by the formula: $E_{n} = -13.6 	imes \frac{Z^2}{n^2} \ 	ext{eV}$.For $Li^{2+}$, the atomic nu

Vedclass · Accepted Answer

$Li^{2+}(n=3), Be^{3+}(n=4)$

Vedclass · Answer

(B) For a single-electron species, the energy is given by the formula: $E_{n} = -13.6 	imes \frac{Z^2}{n^2} \ 	ext{eV}$.For $Li^{2+}$, the atomic number $Z = 3$. At $n = 3$, the energy is $E_3 = -13.6 	imes \frac{3^2}{3^2} = -13.6 \ 	ext{eV}$.For $Be^{3+}$, the atomic number $Z = 4$. At $n = 4$, the energy is $E_4 = -13.6 	imes \frac{4^2}{4^2} = -13.6 \ 	ext{eV}$.Since both species have the same energy of $-13.6 \ 	ext{eV}$, the correct pair is $Li^{2+}(n=3)$ and $Be^{3+}(n=4)$.

Identify the pair of species having the same energy from the following (The number given in the bracket corresponds to the principal quantum number $(n)$ in which the electron is present).

Similar Questions

When an electron transitions from the $6^{th}$ orbit to the $2^{nd}$ orbit in a $H$ atom,what is the total number of spectral lines (excluding the Balmer series) produced?

$A$ metal having a work function of $4 \, eV$ is exposed to a photon of $\lambda = 1240 \, \mathring{A}$. If an accelerating potential of $7.6 \, V$ is applied,then the electron with maximum kinetic energy will have a speed equal to -

The wavenumber associated with the maximum energy in the Balmer series will be......

Which of the following activities can be compared to the concept of quantisation?

When an electron transitions from the $n^{th}$ orbit to the $2^{nd}$ orbit in a $Li^{2+}$ ion,$15$ spectral lines are obtained. What is the value of $n$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module