JEE Main 2024 Chemistry question
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)
Answer: 34
Solution
KE = −E = 3.4 eV = 34 × 10⁻¹ eV.