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Question: If 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron, then the or...

If 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron, then the orbital radius of electron in a hydrogen atom is

A

5.3×10−11 m5.3 \times 10^{- 11}\text{ m}

B

4.3×10−11 m4.3 \times 10^{- 11}\text{ m}

C

6.3×10−11 m6.3 \times 10^{- 11}\text{ m}

D

7.3×10−11 m7.3 \times 10^{- 11}\text{ m}

Answer

5.3×10−11 m5.3 \times 10^{- 11}\text{ m}

Explanation

Solution

Here,

E=−13.6eV=13.6×1.6×10−19=2.2×10−18JE = - 13.6eV = 13.6 \times 1.6 \times 10^{- 19} = 2.2 \times 10^{- 18}J

E=e28πε0EE = \frac{e^{2}}{8\pi\varepsilon_{0}E}

∴\therefore As orbital radius r=−e28πε0Er = \frac{- e^{2}}{8\pi\varepsilon_{0}E}

=9×109×(1.6×10−19)22×(−2.2×10−18)= \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{2 \times ( - 2.2 \times 10^{- 18})}

=5.3×10−11m= 5.3 \times 10^{- 11}m