JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Modern Physics

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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

\frac{9}{x} \times 1 0^{- 10}

m. The value of the

x

is ____.
(

1 eV = 1.6 \times 1 0^{- 19}

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

Nm²/C² and electronic charge

= 1.6 \times 1 0^{- 19}

Answer: 16

Solution

The ionization energy equals the magnitude of the total energy of the electron,

∣ E ∣ = \frac{k e ^{2}}{2 r} = 12.8

eV, so

r = \frac{k e ^{2}}{2∣ E ∣}

. With

k e^{2} = 9 \times 1 0^{9} \times (1.6 \times 1 0^{- 19})^{2}

and

∣ E ∣ = 12.8 \times 1.6 \times 1 0^{- 19}

r = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2 \times 12.8 \times 1.6 \times 1 0 ^{- 19}} = \frac{9}{16} \times 1 0^{- 10}

m. Comparing with

\frac{9}{x} \times 1 0^{- 10}

m, the value of

x

16

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Modern Physics

Solution

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