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Saturday, 31 October

JEE Main 2023 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 6 April 2023, Shift 2, Physics Q56

Question 56 of 90 in this shift, Physics question 26 of 30, Section B.

Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is 9x×10−10\frac{9}{x}\times10^{-10} m. The value of the xx is : __________. (1 eV = 1.6×10−191.6\times10^{-19} J, 14πϵ0=9×109 Nm2/C2\frac{1}{4\pi\epsilon_0}=9\times10^{9}\,\mathrm{Nm^2/C^2} and electronic charge =1.6×10−19=1.6\times10^{-19} C)

Official answer

16

NTA final key.

Chapter
Atoms