Question

What is the work function of the metal if the light of wavelength $4000\mathring{A}$ generates photoelectrons of velocity $6\times {10}^{5}m{s}^{-1}$?

[Given: 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. $0.9eV$
• B. $4.0eV$
• C. $2.1eV$
• D. $3.1eV$

Answer 1

The correct option is C $2.1eV$

$hv=\varphi +h{v}^{\circ }$

$\frac{1}{2}m{v}^2=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

$hv=\varphi +\frac{1}{2}m{v}^2$

$\varphi =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}-\frac{1}{2}\times 9\times {10}^{-31}\times (6\times {10}^5{)}^2$

$\varphi =3.35\times {10}^{-19}J=\varphi \simeq 2.1eV$.

Hence, option C is correct.