Question

What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of velocity $6\times {10}^{5}m{s}^{-1}$ form it?
(Mass of electron $=9\times {10}^{-31}kg$
Velocity of light $=3\times {10}^{8}m{s}^{-1}$
Planck's constant $=6.626\times {10}^{-34}Js$
Charge of electron $=1.6\times {10}^{-19}Je{V}^{-1}$)

• A. 0.9 eV
• B. 4.0 eV
• C. 2.1 eV
• D. 3.1 eV

Answer 1

The correct option is C 2.1 eV

$h\nu =\varphi +\frac{1}{2}m{v}^2$
where,
$h$ is Planck's constant
$\nu$ is frequency of light
$\varphi$ is work function
$m$ is mass of electron
$v$ is velocity of light
$\varphi =h\nu -\frac{1}{2}m{v}^2$
$\varphi =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}-\frac{1}{2}\times 9\times {10}^{-31}\times (6\times {10}^5{)}^2$
$\varphi =3.35\times {10}^{-19}J$
1 eV = $1.6\times {10}^{-19}$ J
~$\Rightarrow \varphi \simeq 2.1eV$
Option C is the correct answer