Question

Light of frequency $8\times 10^{15}Hz$ is incident on a substance of photoelectric work function $6.125eV$. The maximum kinetic energy of the emitted photoelectrons will be.

• A. 39 eV
• B. 27 eV
• C. 54 eV
• D. 13.5 eV

Answer 1

Answer: (B)

From Einstein's photoelectric equation, if the kinetic energy of photoelectrons emitted from metal surface is $E_K$, and $W$ is the work function of metal.
Then $E_K=hv-W$
Where $hv$ is energy of photon absorbed by the electron in the metal.
Given, $v=8\times 10^{15}Hz,$
$h=6.6\times 10^{-34}J-s$
Also $1eV=1.6\times 10^{-19}J$
$W=6.125\times 1.6\times 10^{-19}V$
$=9.8\times 10^{-19}V$
$\therefore E_K=6.6\times 10^{-34}\times 8\times 10^{15}-9.8\times 10^{-19}$
$=52.8\times 10^{-19}-9.8\times 10^{-19}$
$=43\times 10^{-19}J$
Also, $1eV=1.6\times 10^{-19}J$
$\therefore E_K=\frac{43\times 10^{-19}}{1.6\times 10^{-19}}=26.87eV$
$E_K\approx 27eV$
Note: when photon falls on a metal, it transfers whole of its energy to one of the electrons present and its own existence is vanished. The electrons emitted from the surface of a metal have maximum kinetic energy, because their energy is not lost by collisions.