Question

The work function of a metal is $2.5eV$. If the radiation of wavelength $4000A^o$ falls on it, then energy of emitted photoelectrons will be :

• A. $0.402eV$
• B. $0.204eV$
• C. $0.306eV$
• D. $0.603eV$

Answer 1

Answer: (D)

$0.603eV$

$E=hv=\frac{hc}{\lambda }$
$=\frac{6\cdot 63\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-10}}$
$=4\times 966\times {10}^{-19}J$
$=\frac{4\cdot 966\times {10}^{-19}}{1\cdot 6\times {10}^{-19}}eV=3\cdot 103eV$
Energy of emitted photon $=(3\cdot 103-2\cdot 5)eV$
$=0\cdot 603eV$