Kinetic Energy of Electron in First Excited State | JEE Mains PYQ | Structure of Atom
Q. 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)
Solution
Step 1: Total energy relation

For hydrogen atom:
Total Energy (E) = − Kinetic Energy (K.E.)

⇒ K.E. = −E

Step 2: First excited state

First excited state ⇒ n = 2

Given:
E = −3.4 eV

Step 3: Calculate kinetic energy

K.E. = − (−3.4)
K.E. = 3.4 eV

Step 4: Convert into required form

3.4 eV = 34 × 10−1 eV

Final Answer = 34

Previous Year Questions – Structure of Atom
Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is ______ × 10¹³ Hz.
Given below are two statements related to Heisenberg uncertainty principle. Choose the correct answer.
If a₀ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength of the electron present in the second orbit of hydrogen atom?
Radius of the first excited state of Helium ion is given as : a₀ → radius of first stationary state of hydrogen atom.
Which of the following is/are not correct with respect to energy of atomic orbitals of hydrogen atom?
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