What is the maximum velocity of a photoelectron emitted from a surface with work function 5.00 eV when illuminated by 200 nm ultraviolet light? (m_electron = 9.11 x 10^-31 kg,h = 6.63 x 10^-34 J*s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9m) A)800 000 m/s B)653 000 m/s C)431 000 m/s D)212 000 m/s

Question

Quizplus · Accepted Answer

The formula for maximum velocity of a photoelectron is given by $v_{max}=\sqrt{\dfrac{2(E-E_F)}{m}}$ where $E$ is the energy of incident photon, $E_F$ is the work function, and $m$ is the mass of electron.

Converting the given wavelength to energy: $E=\dfrac{hc}{\lambda}=\dfrac{(6.63	imes10^{-34}	ext{ Js})(3	imes10^8	ext{ m/s})}{(200	imes10^{-9}	ext{ m})}=9.94	imes10^{-19}	ext{ J}=6.21	ext{ eV}$

Substituting the given values: 
$v_{max}=\sqrt{\dfrac{2(E-E_F)}{m}}=\sqrt{\dfrac{2(6.21-5.00)	ext{ eV}}{9.11	imes10^{-31}	ext{ kg}}}=653,032	ext{ m/s}\approx \boxed{	extbf{(B)}}$

What Is the Maximum Velocity of a Photoelectron Emitted from a Surface