Question

A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?

Answer 1

The minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron is $1.04 \times 10^{6} \mathrm{ms}-1$

Explanation:

Step 1:

Given data in the question  

Cesium plate Work function, φ = 1.9 eV  

Radiation Wavelength, λ = 250 nm

Photon Energy  

$E=\frac{h c}{\lambda}$

where h = Planck's constant            

           c =Speed of light  

       $\therefore E=\frac{1240}{250}=4.96 e V$

The kinetic energy of an electron, from Einstein's photoelectric equation

 $K=E-\phi$

$K=\frac{h c}{\lambda}-\phi$    

[Here, h is the constant for Planck and c is the speed of light]  

  =3.06 eV.  

For each photoelectron's nonpositive velocity, a photoelectron's velocity should be equivalent to the plate's minimum velocity.

∴ Photoelectron Velocity,

$v=\sqrt{\frac{2 K}{m}}(m=\text { mass of electron })$

$\begin{aligned}v &=\frac{2 \times 3.06 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \\&=1.04 \times 10^{6} \mathrm{ms}-1\end{aligned}$