Question

the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 volt . the photoelectric effect begins in this metal at frequency of 6×10^(14) sec^(-1) . what is the work function of the metal.

Answer 1

Hey dear,

● Answer-

W0 = 2.5 eV

● Explaination-

# Given-

υ0 = 6×10^14 /s

V0 = 3 V

# Solution-

Photoelectric work function of the metal is calculated by-

W0 = hυ0

W0 = 6.63×10^-34 × 6×10^14

W0 = 4×10^-19 V

In terms of electron volts-

W0 = 4×10^-19 / 1.6×10^-19

W0 = 2.5 eV

Therefore, work function of the metal is 2.5 eV.

Hope this is useful...

Answer 2

"Generally, the Kinetic energy is maximum for the ejected electrons. Here in the photoelectric effect the electrons are ejected from the metal surface when light shines on it.

Given,

The frequency of the metal, $υ_0 = 6\times10^{14} /sec$

The retarding potential of light when electron ejects, $V_0 = 3 V$

Photoelectric work function of the metal, $W0 = hυ_0$

We know that h is the Planck’s constant, $h=6.63\times10^{-34}$

$W_0 = 6.63\times10^{-34} \times6\times10^{14}$

$W_0 = 4\times10^{-19} V$

The work function in term of electron volts is given as,

$W_0 = \frac { 4\times10^{ -19 }}{ 1.6\times10^{ -19 } }$

$W_0 = 2.5 eV$

Therefore, the photoelectric work function of the metal is 2.5 eV."