Question

What is the work function of the metal if a light of wavelength $4000\stackrel{\circ }{A}$ generates photoelectrons of velocity $6\times {10}^5{\text{ms}}^{-1}$ from it?
(Mass of electron = $9\times {10}^{-31}\text{kg}$, Velocity of light = $3\times {10}^8{\text{ms}}^{-1}$, Plank's constant = $6.626\times {10}^{-34}\text{Js}$, Charge of electron = $1.6\times {10}^{-19}{\text{J eV}}^{-1}$)

• A. $3.1\text{eV}$
• B. $0.9\text{eV}$
• C. $4.0\text{eV}$
• D. $2.1\text{eV}$

Answer 1

The correct option is D $2.1\text{eV}$

Given:
$\lambda =4000\stackrel{\circ }{A}$
$\nu =6\times {10}^5{\text{ms}}^{-1}$
As, $h\nu =W+K.E$
$\Rightarrow \frac{hc}{\lambda }=W+K.E$

$\Rightarrow \frac{(6.626\times {10}^{-34}\text{Js})\times (3\times {10}^8\text{m/s})}{(4000\times {10}^{-10})\text{m}}=W+\frac{1}{2}m{\text{v}}^2$

$\Rightarrow (5\times {10}^{-19})\text{J}=W+\frac{1}{2}\times (9\times {10}^{-31}\text{kg})\times (6\times {10}^5\text{m/s}{)}^2$

$\Rightarrow W=(5\times {10}^{-19})\text{J}-(1.62\times {10}^{-19})\text{J}$

$\Rightarrow W=3.38\times {10}^{-19}\text{J}$

$\Rightarrow W=\frac{3.38\times {10}^{-19}\text{J}}{1.6\times {10}^{-19}\text{J/eV}}$

$\Rightarrow W=2.11\text{eV}$