Question

A) The work function of caesium metal is $2.14eV$. When light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. What is the maximum kinetic energy of the emitted electrons?

B) The work function of caesium metal is $2.14eV$. When light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. Maximum kinetic energy of the photoelectron is $0.34ev$. What is the stopping potential of the emitted electrons?

C) The work function of caesium metal is $2.14eV$. When light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. Maximum kinetic energy of the photoelectron is $0.345ev$. What is the of the maximum speed of the emitted photoelectrons?

Answer 1

A) Given,
Work function of cesium metal, $W_0=2.14eV$
Frequency of incident light, $v=6\times {10}^{14}Hz$
The maximum kinetic energy of the photoelectron
$K_{max}=hv-W_0$
Where,
$h$ = Plnack's constant
$h=6.626\times {10}^{-34}Js$
Therefore,
$K_{max}=\frac{6.626\times {10}^{-34}\times 6\times {10}^{14}}{1.6\times {10}^{-19}}-2.14$
$K_{max}=2.485-2.14$
$K_{max}=0.345eV$
As we know, $1ev=1.6\times {10}^{-19}J$
Therefore,
$K_{max}=0.345\times 1.6\times {10}^{-19}J$
$K_{max}=5.52\times {10}^{-20}J$
Hence, the maximum kinetic energy of the emitted electrons is 0$.345eV$.
Final Answer : $K_{max}=0.345eV$.

B) The maximum kinetic energy of photoelectrons is related to stopping potential, $V_0$
$K_{max}=e{V}_0$
$V_0=\frac{K_{max}}{e}$
$V_0=\frac{0.34eV}{e}$
As we know, $1eV=1.6\times {10}^{-19}J$
Therefore,
$V_0=\frac{0.34\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}$
$V_0=0.34V$
Hence, the stopping potential of the material is 0$.34V$.
Final Answer: $V_0=0.34V$.

C) Maximum kinetic energy is given by,
$K_{max}=\frac{1}{2}m{v}^2$
Maximum speed of the emitted photoelectrons,
$v=\sqrt{\frac{2K_{max}}{m}}$
Where,
$m$ = Mass of an electron
$m=9.1\times {10}^{-31}\text{kg}$
As we know,
$1ev=1.6\times {10}^{-19}J$
Therefore,
$K_{max}=0.345\times 1.6\times {10}^{-19}J$
$K_{max}=5.52\times {10}^{-20}J$
Maximum speed of the emitted photoelectrons,
$v=\sqrt{\frac{2K_{max}}{m}}$
$v=\sqrt{\frac{2\times 5.52\times {10}^{-20}}{9.11\times {10}^{-31}}}$
$v=348000m/s$
$v=348\text{km/s}$
Final Answer: $v=348\text{km/s}$.