Question

Calculate the maximum KE and velocity of an electron from zinc by a 275nm photon

Answer 1

Answer:

Kinetic energy: $6.024*10^{-20}$

Velocity: $3.64*10^{5}$

Explanation:

From the equations of the photo-electric effect,

We know:

$hv=hv_0+K.E$

Where,

1.$h$ is the Planck's constant which is $6.626*10^{-34}$

2.$v,v_0$ are the frequency of light emitted and threshold frequencies respectively.

3.$K.E$ is the kinetic energy of the electrons emitted.

By fact, we come to know that the threshold frequency of Zn is 300nm

And also $v=\frac{c}{d}$

Where ,

1. $c$ is the speed of light $=3*10^8$

2.$d$ is the wavelength.

Thus,

$\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}$

Now to find velocity:

$K.E=\frac{1}{2}mv^2\\v^2=1.324*10^{11}\\v=3.64*10^5$