Question

When electromagnetic radiation of wavelength 300 nm 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 minimum energy needed to remove an electron from sodium?

• A. $E_{min}=3.84\times {10}^{-19}J$
• B. $E_{min}=2.31\times {10}^{5}J$
• C. $E_{min}=4.42\times {10}^{-19}J$
• D. $E_{min}=3.78\times {10}^{-19}J$

Answer 1

The correct option is A $E_{min}=3.84\times {10}^{-19}J$

The energy (E) of a 300 nm photon is given by
$hv=\frac{hc}{\lambda }\because \lambda \nu =c$
$=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}{300\times {10}^{-9}m}$
$=6.626\times {10}^{-19}J$
$\text{The energy of one mole of photons}=6.626\times {10}^{-19}J\times 6.022\times {10}^{23}{\text{mol}}^{-1}=3.99\times {10}^5{\text{Jmol}}^{-1}$
$\text{The minimum energy needed to remove a mole of electrons from sodium = Energy of photons - Kinetic energy of electrons}$
$=(3.99-1.68)\times {10}^5{\text{Jmol}}^{-1}$
$=2.31\times {10}^5{\text{Jmol}}^{-1}$
The minimum energy for one electron $=\frac{2.31\times {10}^5{\text{Jmol}}^{-1}}{6.022\times {10}^{23}{\text{electrons mol}}^{-1}}$
$=3.84\times {10}^{-19}J$