Q87 / 90 · JEE Main 2023 (06 Apr Shift 2) · Physics

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 m.

The value of the x is: _____.

(1 eV=1.6×10-19 J, 14πε0=9×109 Nm2C2 and electronic charge =1.6×10-19 C)

Correct answer: 16

Solution

The given data is 

q=1.6×10-19 C14πε0=9×109 N m2 C-2

The formula for the value of energy in a hydrogen atom is given by 

E=q28πε0r⇒E=q22(4πε0)r

⇒r=9×109×1.6×10-1912.8×2

⇒r=916×10-10 m

It is given that r=9x×10-10 m⇒916×10-10=9x×10-10 ⇒x=16

Hence, the correct answer is 16.

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