Question

The work function of cesium metal is $2.14ev.$ When the light of frequency $6\ast {10}^{1}4$Hz is incident on the metal surface photoemission of electrons occurs.What is the maximum kinetic energy of the emitted electrons.

• A. $0.35eV.$
  
• B. $0.34eV$
  
• C. $0.29eV$
  
• D. $0.13eV$
  

Answer 1

The correct option is A $0.35eV.$

Given,

$\varphi =2.14eV$

$f=6\times {10}^{14}Hz$

The energy of photon,

$E=hf=\frac{6.6\times {10}^{-34}\times 6\times {10}^{14}}{1.6\times {10}^{-19}}eV$

$E=2.485eV$

From Einstein photoelectric equation,

$E=K_{max}+\varphi$

$K_{max}=E-\varphi =2.485eV-2.14eV$

$K_{max}=0.35eV$

The correct option is A.