Question

The velocity of electron in hydrogen atom is ${7.29\times 10}^7$ cm/sec. The potential energy of that electron is:

• A. $-13.6eV$
• B. $-3.4eV$
• C. $-3.02eV$
• D. $-1.70eV$

Answer 1

Answer: (B)

$-3.4eV$

Velocity of electron in $n^{th}$ orbit is $2.165\times {10}^6\frac{z}{n}$${ms}^{-1}$.

given, velocity of electron is $7.29\times {10}^7\times {10}^{-2}m/sec$

$7.29\times {10}^5\frac{m}{sec}=2.165\times {10}^6\times \frac{1}{n}m/sec$

$n\cong 3$

$\Rightarrow$ electron resolving in $3^{rd}$ orbit.

Potential energy of that electron is

$P_E=\frac{{-ze}^2}{4\Pi {\epsilon }_0{\pi }_n}=\frac{(-1){(1.6\times {10}^{-19})}^2\left({10}^{10}\right)}{4\Pi (885\times {10}^{-12})(0.512\times 3^2)}$

potential energy of that electron $=3.4ev$