Match the following items:A. Energy of ground | vedclass

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

Question

(C) For $He^+$ $(Z=2)$:$(A)$ Ground state energy $(n=1)$: $E_n = -13.6 	imes (Z^2/n^2) 	ext{ eV} = -13.6 	imes (4/1) = -54.4 	ext{ eV}$. Thus,$A-i

Vedclass · Accepted Answer

$A-iv, B-ii, C-i, D-iii$

Vedclass · Answer

(C) For $He^+$ $(Z=2)$:$(A)$ Ground state energy $(n=1)$: $E_n = -13.6 	imes (Z^2/n^2) 	ext{ eV} = -13.6 	imes (4/1) = -54.4 	ext{ eV}$. Thus,$A-iv$.$(B)$ Potential energy of $H$ atom $(Z=1)$ in $I$ orbit $(n=1)$: $PE = 2 	imes E_n = 2 	imes (-13.6 	ext{ eV}) = -27.2 	ext{ eV}$. Thus,$B-ii$.$(C)$ Kinetic energy of $II$ excited state $(n=3)$ of $He^+$: $KE = |E_n| = 13.6 	imes (Z^2/n^2) = 13.6 	imes (4/9) \approx 6.04 	ext{ eV}$. Thus,$C-i$.$(D)$ Ionization potential of $He^+$: Energy required to remove electron from $n=1$ to $\infty$,which is $54.4 	ext{ eV}$. Converting to Joules: $54.4 	imes 1.602 	imes 10^{-19} 	ext{ J} \approx 8.72 	imes 10^{-18} 	ext{ J}$. Thus,$D-iii$.Therefore,the correct match is $A-iv, B-ii, C-i, D-iii$.

$A$. Energy of ground state of $He^+$	$i$. $+6.04 \text{ eV}$
$B$. Potential energy of $I$ orbit of $H$ atom	$ii$. $-27.2 \text{ eV}$
$C$. Kinetic energy of $II$ excited state of $He^+$	$iii$. $8.72 \times 10^{-18} \text{ J}$
$D$. Ionization potential of $He^+$	$iv$. $-54.4 \text{ eV}$

Similar Questions

Which of the following graphs represents one node?

Which of the following arrangements of electrons is most likely to be stable?

Which of the following graphs is correct for the kinetic energy $(K.E.)$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom?

Which of the following violates the Aufbau principle and Hund's rule?

Vedclass Test Series

Exam Paper Generator

Online Exam Module