Question

In an experiment, a student found that a maximum wavelength of 351 nm is needed to just dislodge electrons from a metal surface. calculate the velocity (in m/s) of an ejected electron when the student employed light with a wavelength of 313 nm.

Answer 1

Before solving this question first we have to understand work function.

The work function of a metal is amount of minimum energy required to emit an electron from the surface barrier of metal . Whenever the metal will be exposed to radiation a part of its energy will be utilized  to emit an electron while rest will provide kinetic energy to the electron.

Let  f is the frequency of incident radiation and f' is the frequency corresponding to work function. Let v is the velocity of the ejected electron.

we know that velocity of an electromagnetic wave is the product of frequency and wavelength. Hence frequency f is given as-

                                       $f=\frac{c}{\lambda}$

where c is velocity of light and $\lambda$ is the wavelength of the wave.

As per the question incident wavelength =313 nm

                                                                     $313*10^{-9} m$  [as 1 nm =10^-9 m]

The wavelength corresponding to work function is 351 nm i.e

                                                                           $351*10^{-9} m$

                      we know that  hf=hf'+K.E [ h is the planck's constant whose value is 6.63×10^-34 J-s]

                                          ⇒K.E =hf-hf'

                                           $\frac{1}{2} mv^2=hf-hf'$

                                            $v^2=\frac{2}{m} [hf-hf']$

                                     $v^2=\frac{2}{m} [\frac{hc}{\lambda} -\frac{hc}{\lambda ' }]$

                                                                       $=\frac{2}{9.1*10^{-31}kg } *{6.63*10^{-34} Js *3*10^{8} [\frac{1}{313*10^{-9} } -\frac{1}{351*10^{-9} } ]$

                                           $=0.001512021301356*10^{14} m^2/s^2$

                                           $v=\sqrt{0.0015120241301356*10^{14} } m/s$

                                                 $=0.3888476161313*10^{7} m/s$

                                                 $=3.88848*10^{7} m/s$  [ans]

Answer 2

Answer:

$v = 3.94 \times 10^5 m/s$

Explanation:

As we know that the maximum wavelength needed to remove the electron from metal surface is given as

$\lambda = 351 nm$

so we have work function of metal is given as

$\phi = \frac{hc}{\lambda}$

$\phi = \frac{1240}{351} eV$

$\phi = 3.53 eV$

Now as per Einstein's equation for photo electric effect we know that

$E = \phi + \frac{1}{2}mv^2$

$\frac{hc}{\lambda} = \phi + \frac{1}{2}mv^2$

$(\frac{1240}{313} - 3.52)1.6 \times 10^{-19} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$7.07 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 3.94 \times 10^5 m/s$