Question

When electromagnetic radiation of wavelength $300nm$ falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68\times {10}^{5}Jmo{l}^{-1}$. What is the maximum wavelength that will cause a photoelectron to be emitted?

• A. $\lambda =320nm$
• B. $\lambda =449.7nm$
• C. $\lambda =526nm$
• D. $\lambda =518nm$

Answer 1

The correct option is D $\lambda =518nm$

The energy of a $300nm$ photon $=hv=\frac{hc}{\lambda }$
$=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}{300\times {10}^{-9}m}$
$=6.626\times {10}^{-19}J$

The energy of one mole of photon $=6.626\times {10}^{-19}J\times 6.022\times {10}^{23}mo{l}^{-1}$
$=3.99\times {10}^{5}Jmo{l}^{-1}$

$\text{The minimum energy needed to remove a mole of electrons from sodium}=\text{Energy of photons - Kinetic energy of electrons}$
$=(3.99-1.68)\times {10}^{5}Jmo{l}^{-1}$
$=2.31\times {10}^{5}Jmo{l}^{-1}$

The minimum energy for one electron $=\frac{2.31\times {10}^{5}Jmo{l}^{-1}}{6.022\times {10}^{23}mo{l}^{-1}}$
$=3.84\times {10}^{-19}J$

$\therefore \lambda =\frac{hc}{E}$
$=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}{3.84\times {10}^{-19}J}$
$=5.18\times {10}^{-7}m=518nm$