For hydrogen atom, energy of an electron in first excited state is (-3.4 eV, K . E ). of the same electron of hydrogen atom is (x eV ). Value of x is ______ ( × 10^ -1 eV ). (Nearest integer)

Correct answer: 0 Kinetic energy of electron is, KE= 13 · 6 Z^2 n^2 eV For the first excited state of the hydrogen atom, n=2 and Z=1 KE=(13.6)/(2^2)=3.4 eV

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Structure of Atom

JEE Main 2024 · 6 April · Evening Shift · Chemistry · Structure of Atom

For hydrogen atom, energy of an electron in first excited state is...

For hydrogen atom, energy of an electron in first excited state is (-3.4

\mathrm{eV}, \mathrm{K} . \mathrm{E}

). of the same electron of hydrogen atom is (x

\mathrm{eV}

). Value of

x

is ______ (

\times

10^{-1}

\mathrm{eV}

). (Nearest integer)

Correct answer

Step-by-step solution

Kinetic energy of electron is,

\mathrm{KE}=\frac{13 \cdot 6 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{eV}

For the first excited state of the hydrogen atom,

\mathrm{n}=2

and

\mathrm{Z}=1

\therefore \mathrm{KE}=\frac{13.6}{2^2}=3.4 \mathrm{eV}

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