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).

Chemistry · Structure of Atom

JEE Main 2024 — 6 April, Shift 2 — Question 85

For hydrogen atom, energy of an electron in first excited state is $-3.4\ \mathrm{eV}$ . K.E. of the same electron of hydrogen atom is $x\ \mathrm{eV}$ . Value of $x$ is $\_\_\_\_\_$$\times 10^{-1}\ \mathrm{eV}$ (nearest integer).

Answer: 34

Numerical answer — enter this value.

Step-by-step solution

For Bohr atom, $\mathrm{E = -K.E.}$

Given:

\mathrm{E = -3.4\ eV}

\mathrm{K.E. = 3.4\ eV}

Expressing in required form, we have

\mathrm{K.E. = 34 \times 10^{-1}\ eV}

\mathrm{x = 34}

Thus, the value of $x$ is $34$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Chemistry
Chapter: Structure of Atom
Topic: Bohr's Model of Atom