Question

The energy required to separate a hydrogen atom into a proton and an electron is 13.6eV. Then the velocity of electron in a hydrogen atom is

• A. ${2.2\times 10}^{4}m/s$
• B. ${2.2\times 10}^{2}m/s$
• C. ${2.2\times 10}^{6}m/s$
• D. ${2.2\times 10}^{10}m/s$

Answer 1

The correct option is C ${2.2\times 10}^{6}m/s$

Given,

$E=13.6eV$

$m_e=9.1\times {10}^{-31}kg$

The kinetic energy is equal to $13.6eV$,

$\frac{1}{2}{m}_e{v}^2=13.6eV$

$v^2=\frac{13.6eV\times 2}{m_e}$

$v^2=\frac{13.6\times 1.6\times {10}^{-19}\times 2}{9.1\times {10}^{-31}}=4.78\times {10}^{12}$

$v=\sqrt{4.78\times {10}^{12}}m/s$

$v=2.2\times {10}^{6}m/s$

The correct option is C.