Question

Light of frequency $8 \times 10^{15} \,Hz$ is incident on a substance of photoelectric work function $6.125 \,eV$. 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 $E_{K}=h v-W$
Where $h v$ is energy of photon absorbed by the electron in the metal.
Given, $v=8 \times 10^{15} \,H z, $
$h=6.6 \times 10^{-34} \,J-s$
Also $1 \,e V=1.6 \times 10^{-19} \,J$
$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, $ 1\, e V=1.6 \times 10^{-19} J $
$\therefore E_{K}=\frac{43 \times 10^{-19}}{1.6 \times 10^{-19}}=26.87 \,eV $
$E_{K} \approx 27\, eV$
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.