Question

When electromagnetic radiation of wavelength $300\, nm$ falls on the surface of a metal, electrons are emitted with the kinetic energy of $1.68 \times 10^5\, J\, mol ^{-1}$. What is the minimum energy needed to remove an electron from the metal?
$\left( h =6.626 \times 10^{-34} Js , c =3 \times 10^8 ms ^{-1}\right. \text {, }\left.N _{ A }=6.022 \times 10^{23} \, mol ^{-1}\right)$

• A. $2.31 \times 10^6\, J\, mol ^{-1}$
• B. $3.84 \times 10^4 \, J \, mol ^{-1}$
• C. $3.84 \times 10^{-19} \, J \, mol ^{-1}$
• D. $2.31 \times 10^5\, J \, mol ^{-1}$

Answer 1

Answer: (D)

Energy of a $300\, nm$ photon is given by $\Rightarrow$
$E =\frac{ hc }{\lambda}=\frac{6.626 \times 10^{-34} \, Js \times 3 \times 10^8 ms ^{-1}}{300 \times 10^{-9}\, m } $
$=6.626 \times 10^{-19} J$
Energy of one mole of photons
$=6.626 \times 10^{-19} J \times 6.022 \times 10^{23} \, mol ^{-1}$
$=3.99 \times 10^5 \, J\, mol ^{-1}$
The minimum energy needed to remove one mole of electrons $=(3.99-1.68) \times 10^5 \, J\, mol ^{-1}$
$=2.31 \times 10^5 \, J \, mol ^{-1}$