Question

What is the maximum velocity (in m/s) of electrons ejected from a material by 66 nm photons if the work function of the material is 4.43 eV?

Answer 1

$h$ - Plank constant

$\nu$ - frequency of photon

$\lambda$ - wave lenght of photon

$c$ - speed of light

$A$ - work function of the material

$m_e$ - mass of electron

$V_m$ - the maximum velocity of electrons ejected from a material

$\dfrac{c}{\nu}=\lambda \Rightarrow \nu=\dfrac{c}{\lambda}$

Energy of photon $h\nu=\dfrac{hc}{\lambda}$ (according to hypothesis of Plank)

According to energy Einstein formula

$h\nu=\dfrac{m_eV_m^2}{2}+A \Rightarrow \dfrac{hc}{\lambda} = \dfrac{m_eV_m^2}{2}+A$

$\dfrac{hc}{\lambda} -A= \dfrac{m_eV_m^2}{2}$

$\dfrac{2hc}{\lambda} -2A= m_eV_m^2$

$\dfrac{2hc}{\lambda} -\dfrac{2A\lambda}{\lambda}= m_eV_m^2$

$\dfrac{2hc-2A\lambda}{\lambda} = m_eV_m^2$

$\dfrac{2(hc-A\lambda)}{\lambda} = m_eV_m^2$

$\dfrac{2(hc-A\lambda)}{\lambda m_e} = V_m^2$

$V_m=\sqrt{\dfrac{2(hc-A\lambda)}{\lambda m_e}}\approx2.25*10^6 \dfrac{m}{s}$

Answer: $2.25*10^6 \dfrac{m}{s}$