Question

The work function of cesium metal is $2.14eV$. When the light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. What is the

1. the maximum kinetic energy of the emitted electrons
2. Stopping potential
3. the maximum speed of the emitted photoelectrons

Answer 1

Step 1: Given Data

The work function of Caesium metal is $2.14$eV

Light of frequency $6\times 1014$ Hz

Step 2: Solve for the maximum Kinetic Energy of the emitted electrons

As we know that

$K.E.=h\nu e-\varphi$

$\nu =$frequency of light

$h=$Planck's constant$=6.626\times {10}^{-34}$

$e=$charge of an electron$=1.6\times {10}^{-19}$

$\varphi =$the work function of Caesium

On substituting the known values in the equation,

$$\begin{gathered} \Rightarrow K.E.=6.626\times {10}^{-34}\times 6\times 1014\times 1.6\times {10}^{-19}-2.14 \\ \Rightarrow K.E.=2.485-2.14 \\ \Rightarrow K.E.=0.345eV \end{gathered}$$

Therefore, the maximum kinetic energy of the emitted electrons is$0.345eV$.

Step 3: Solve for stopping potential

As we know

Stopping potential $V_{\circ }=\frac{K.E.}{e}$

$$\begin{gathered} V_{\circ }=\frac{0.345\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}} \\ {V}_{\circ }=0.345 \end{gathered}$$

Therefore stopping potential is $0.345V$

Step 4: Maximum speed of the emitted photoelectrons

As we know that

$K.E.=\frac{1}{2}m{v}^2$, where

$m=$mass of an electron$=9.1\times {10}^{-31}$

$v=$speed of the electron

$$\begin{gathered} \Rightarrow v=\sqrt{\frac{2K.E.}{m}} \\ \Rightarrow v=\sqrt{\frac{2\times 0.345\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}} \\ \Rightarrow v=348.3km/sec \end{gathered}$$

Therefore, the maximum speed of the emitted photoelectrons is $348.3km/sec$.

Hence,

The maximum kinetic energy of the emitted electrons is $0.345eV$.

Stopping potential is $0.345V$.

the maximum speed of the emitted photoelectrons is $348.3km/sec$.