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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2023 · 6 April · Evening Shift · Physics · Electrostatics

Experimentally it is found that 12 . 8eV energy is required to...

Experimentally it is found that 12.8eV12 . 8eV 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×10−10m\dfrac{9}{x} \times {10}^{- 10}m.

The value of the xx is: _____.

(1eV=1.6×10−19J,14πε0=9×109Nm2C21eV = 1 . 6 \times {10}^{- 19}J, \dfrac{1}{4\pi{ε}_{0}} = 9 \times {10}^{9}\dfrac{{Nm}^{2}}{{C}^{2}} and electronic charge =1.6×10−19C= 1 . 6 \times {10}^{- 19}C)

Correct answer

0

Step-by-step solution

The given data is

q=1.6×10−19C14πε0=9×109Nm2C−2q = 1 . 6 \times {10}^{- 19}C\dfrac{1}{4{\pi ε}_{0}} = 9 \times {10}^{9}N{m}^{2}{C}^{- 2}

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

E=q28πε0r⇒E=q22(4πε0)rE = \dfrac{{q}^{2}}{8\pi{ε}_{0}r} ⇒ E = \dfrac{{q}^{2}}{2 ( 4\pi{ε}_{0} ) r}

⇒r=9×109×1.6×10−1912.8×2⇒ r = \dfrac{9 \times {10}^{9} \times 1 . 6 \times {10}^{- 19}}{12 . 8 \times 2}

⇒r=916×10−10m⇒ r = \dfrac{9}{16} \times {10}^{- 10}m

It is given that r=9x×10−10m⇒916×10−10=9x×10−10⇒x=16r = \dfrac{9}{x} \times {10}^{- 10}m ⇒ \dfrac{9}{16} \times {10}^{- 10} = \dfrac{9}{x} \times {10}^{- 10} ⇒ x = 16

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